Bài 31 toán 8 tập 2 trang 23 năm 2024

Bài 31 Trang 23 SGK Toán 8 tập 2 biên soạn và đăng tải với hướng dẫn chi tiết lời giải giúp cho các em học sinh tham khảo, ôn tập, củng cố kỹ năng giải Toán 8. Mời các em học sinh cùng tham khảo chi tiết.

Bài 31 Trang 23 SGK Toán 8 - Tập 2

Bài 31 (SGK trang 23): Giải các phương trình:

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%5Cleft(%202x%2B7%20%5Cright)%7D%2B%5Cfrac%7B1%7D%7B2x%2B7%7D%3D%5Cfrac%7B6%7D%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B3%20%5Cright)%7D)

Hướng dẫn giải

Bước 1: Đặt điều kiện xác định (ĐKXĐ) của phương trình.

Bước 2: Quy đồng khử mẫu

Bước 3: Sử dụng quy tắc chuyển vế để tìm a.

Bước 4: Kiểm tra giá trị của a tìm được có thỏa mãn với ĐKXĐ

Bước 5: Kết luận.

Lời giải chi tiết

Điều kiện xác định: ![\left{ \begin{matrix} x-1\ne 0 \ {{x}^{{3}}-1\ne 0 \
{{x}}{2}}+x+1\ne 0 \ \end{matrix} \right.\Leftrightarrow \left{ \begin{matrix} x-1\ne 0 \ \left( x-1 \right)\left( {{x}^{{2}}+x+1 \right)\ne 0 \
{{x}}{2}}+x+1\ne 0 \ \end{matrix} \right.\Leftrightarrow \left{ \begin{matrix} x-1\ne 0 \ {{x}^{2}}+x+1\ne 0 \ \end{matrix} \right.](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-1%5Cne%200%20%5C%5C%0A%0A%7B%7Bx%7D%5E%7B3%7D%7D-1%5Cne%200%20%5C%5C%0A%0A%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-1%5Cne%200%20%5C%5C%0A%0A%5Cleft(%20x-1%20%5Cright)%5Cleft(%20%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%20%5Cright)%5Cne%200%20%5C%5C%0A%0A%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-1%5Cne%200%20%5C%5C%0A%0A%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.)

![\Leftrightarrow \left{ \begin{matrix} x-1\ne 0 \ {{x}^{{2}}+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\ne 0 \
\end{matrix}\Leftrightarrow \left{ \begin{matrix}
x\ne 1 \
{{\left( x+\dfrac{1}{2} \right)}}{2}}+\dfrac{3}{4}\ge 0\forall x \ \end{matrix} \right. \right.\Leftrightarrow x\ne 1](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-1%5Cne%200%20%5C%5C%0A%0A%7B%7Bx%7D%5E%7B2%7D%7D%2B2.x.%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B1%7D%7B4%7D-%5Cdfrac%7B1%7D%7B4%7D%2B1%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%5Cne%201%20%5C%5C%0A%0A%7B%7B%5Cleft(%20x%2B%5Cdfrac%7B1%7D%7B2%7D%20%5Cright)%7D%5E%7B2%7D%7D%2B%5Cdfrac%7B3%7D%7B4%7D%5Cge%200%5Cforall%20x%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%20%5Cright.%5CLeftrightarrow%20x%5Cne%201)

![\begin{align} & \frac{1}{x-1}-\frac{3{{x}^{2}}}{{{x}{3}}-1}=\frac{2x}{{{x}^{{2}}+x+1} \
& \Leftrightarrow \frac{1}{x-1}-\frac{3{{x}}{2}}}{\left( x-1 \right)\left( {{x}^{{2}}+x+1 \right)}=\frac{2x}{{{x}}{2}}+x+1} \ & \Leftrightarrow \frac{{{x}^{{2}}+x+1}{\left( {{x}}{2}}+x+1 \right)\left( x-1 \right)}-\frac{3{{x}^{{2}}}{\left( x-1 \right)\left( {{x}}{2}}+x+1 \right)}=\frac{2x\left( x-1 \right)}{\left( {{x}^{{2}}+x+1 \right)\left( x-1 \right)} \
& \Leftrightarrow {{x}}{2}}+x+1-3{{x}^{2}}=2{{x}{2}}-2x \ & \Leftrightarrow -2{{x}^{{2}}+x+1=2{{x}}{2}}-2x \ & \Leftrightarrow -4{{x}^{{2}}+3x+1=0 \
& \Leftrightarrow -4{{x}}{2}}+4x-x+1=0 \ & \Leftrightarrow -4x\left( x-1 \right)-\left( x-1 \right)=0 \ & \Leftrightarrow \left( x-1 \right)\left( -4x-1 \right)=0 \ & \Leftrightarrow \left[ \begin{matrix} x-1=0 \ 4x+1=0 \ \end{matrix} \right.\Leftrightarrow \left[ \begin{matrix} x=1\left( L \right) \ x=\frac{-1}{4}\left( tm \right) \ \end{matrix} \right. \ \end{align}](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cbegin%7Balign%7D%0A%0A%26%20%5Cfrac%7B1%7D%7Bx-1%7D-%5Cfrac%7B3%7B%7Bx%7D%5E%7B2%7D%7D%7D%7B%7B%7Bx%7D%5E%7B3%7D%7D-1%7D%3D%5Cfrac%7B2x%7D%7B%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cfrac%7B1%7D%7Bx-1%7D-%5Cfrac%7B3%7B%7Bx%7D%5E%7B2%7D%7D%7D%7B%5Cleft(%20x-1%20%5Cright)%5Cleft(%20%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%20%5Cright)%7D%3D%5Cfrac%7B2x%7D%7B%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cfrac%7B%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%7D%7B%5Cleft(%20%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%20%5Cright)%5Cleft(%20x-1%20%5Cright)%7D-%5Cfrac%7B3%7B%7Bx%7D%5E%7B2%7D%7D%7D%7B%5Cleft(%20x-1%20%5Cright)%5Cleft(%20%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%20%5Cright)%7D%3D%5Cfrac%7B2x%5Cleft(%20x-1%20%5Cright)%7D%7B%5Cleft(%20%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%20%5Cright)%5Cleft(%20x-1%20%5Cright)%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1-3%7B%7Bx%7D%5E%7B2%7D%7D%3D2%7B%7Bx%7D%5E%7B2%7D%7D-2x%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20-2%7B%7Bx%7D%5E%7B2%7D%7D%2Bx%2B1%3D2%7B%7Bx%7D%5E%7B2%7D%7D-2x%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20-4%7B%7Bx%7D%5E%7B2%7D%7D%2B3x%2B1%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20-4%7B%7Bx%7D%5E%7B2%7D%7D%2B4x-x%2B1%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20-4x%5Cleft(%20x-1%20%5Cright)-%5Cleft(%20x-1%20%5Cright)%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cleft(%20x-1%20%5Cright)%5Cleft(%20-4x-1%20%5Cright)%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-1%3D0%20%5C%5C%0A%0A4x%2B1%3D0%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%3D1%5Cleft(%20L%20%5Cright)%20%5C%5C%0A%0Ax%3D%5Cfrac%7B-1%7D%7B4%7D%5Cleft(%20tm%20%5Cright)%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%20%5C%5C%0A%0A%5Cend%7Balign%7D)

Vậy là nghiệm duy nhất của phương trình.

Điều kiện xác định: ![\left{ \begin{matrix} x-1\ne 0 \ x-2\ne 0 \ x-3\ne 0 \ \end{matrix}\Leftrightarrow \left{ \begin{matrix} x\ne 1 \ x\ne 2 \ x\ne 3 \ \end{matrix} \right. \right.](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-1%5Cne%200%20%5C%5C%0A%0Ax-2%5Cne%200%20%5C%5C%0A%0Ax-3%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%5Cne%201%20%5C%5C%0A%0Ax%5Cne%202%20%5C%5C%0A%0Ax%5Cne%203%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%20%5Cright.)

![\begin{align} & \frac{3}{\left( x-1 \right)\left( x-2 \right)}+\frac{2}{\left( x-3 \right)\left( x-1 \right)}=\frac{1}{\left( x-2 \right)\left( x-3 \right)} \ & \Leftrightarrow \frac{3\left( x-3 \right)}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}+\frac{2\left( x-2 \right)}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}=\frac{1\left( x-1 \right)}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)} \ & \Leftrightarrow 3\left( x-3 \right)+2\left( x-2 \right)=x-1 \ & \Leftrightarrow 3x-9+2x-4=x-1 \ & \Leftrightarrow 5x-13=x-1 \ & \Leftrightarrow 4x=12 \ & \Leftrightarrow x=3\left( L \right) \ \end{align}](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cbegin%7Balign%7D%0A%0A%26%20%5Cfrac%7B3%7D%7B%5Cleft(%20x-1%20%5Cright)%5Cleft(%20x-2%20%5Cright)%7D%2B%5Cfrac%7B2%7D%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x-1%20%5Cright)%7D%3D%5Cfrac%7B1%7D%7B%5Cleft(%20x-2%20%5Cright)%5Cleft(%20x-3%20%5Cright)%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cfrac%7B3%5Cleft(%20x-3%20%5Cright)%7D%7B%5Cleft(%20x-1%20%5Cright)%5Cleft(%20x-2%20%5Cright)%5Cleft(%20x-3%20%5Cright)%7D%2B%5Cfrac%7B2%5Cleft(%20x-2%20%5Cright)%7D%7B%5Cleft(%20x-1%20%5Cright)%5Cleft(%20x-2%20%5Cright)%5Cleft(%20x-3%20%5Cright)%7D%3D%5Cfrac%7B1%5Cleft(%20x-1%20%5Cright)%7D%7B%5Cleft(%20x-1%20%5Cright)%5Cleft(%20x-2%20%5Cright)%5Cleft(%20x-3%20%5Cright)%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%203%5Cleft(%20x-3%20%5Cright)%2B2%5Cleft(%20x-2%20%5Cright)%3Dx-1%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%203x-9%2B2x-4%3Dx-1%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%205x-13%3Dx-1%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%204x%3D12%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20x%3D3%5Cleft(%20L%20%5Cright)%20%5C%5C%0A%0A%5Cend%7Balign%7D)

Vậy phương trình vô nghiệm

Điều kiện xác định: ![\left{ \begin{matrix} x+2\ne 0 \ 8+{{x}^{{3}}\ne 0 \
\end{matrix} \right.\Leftrightarrow \left{ \begin{matrix}
x\ne -2 \
\left( x+2 \right)\left( 4-2x+{{x}}{2}} \right)\ne 0 \ \end{matrix} \right.\Leftrightarrow \left{ \begin{matrix} x\ne -2 \ 4-2x+{{x}^{2}}\ne 0 \ \end{matrix} \right.](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%2B2%5Cne%200%20%5C%5C%0A%0A8%2B%7B%7Bx%7D%5E%7B3%7D%7D%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%5Cne%20-2%20%5C%5C%0A%0A%5Cleft(%20x%2B2%20%5Cright)%5Cleft(%204-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%20%5Cright)%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%5Cne%20-2%20%5C%5C%0A%0A4-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.)

![\Leftrightarrow \left{ \begin{matrix} x\ne -2 \ 3+1-2x+{{x}^{{2}}\ne 0 \
\end{matrix} \right.\Leftrightarrow \left{ \begin{matrix}
x\ne -2 \
3+{{\left( 1-x \right)}}{2}}\ne 0\forall x \ \end{matrix} \right.\Rightarrow x\ne -2](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%5Cne%20-2%20%5C%5C%0A%0A3%2B1-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%5Cne%20-2%20%5C%5C%0A%0A3%2B%7B%7B%5Cleft(%201-x%20%5Cright)%7D%5E%7B2%7D%7D%5Cne%200%5Cforall%20x%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CRightarrow%20x%5Cne%20-2)

![\begin{align} & 1+\frac{1}{x+2}=\frac{12}{8+{{x}^{{3}}} \
& \Leftrightarrow 1+\frac{1}{x+2}=\frac{12}{\left( x+2 \right)\left( 4-2x+{{x}}{2}} \right)} \ & \Leftrightarrow \frac{\left( x+2 \right)\left( 4-2x+{{x}^{{2}} \right)}{\left( x+2 \right)\left( 4-2x+{{x}}{2}} \right)}+\frac{4-2x+{{x}^{{2}}}{\left( x+2 \right)\left( 4-2x+{{x}}{2}} \right)}=\frac{12}{\left( x+2 \right)\left( 4-2x+{{x}^{{2}} \right)} \
& \Leftrightarrow 8+{{x}}{3}}+4-2x+{{x}^{{2}}=12 \
& \Leftrightarrow {{x}}{3}}+{{x}^{{2}}-2x=0 \
& \Leftrightarrow x\left( {{x}}{2}}-x+2x-2 \right)=0 \ & \Leftrightarrow x\left[ x\left( x-1 \right)+2\left( x-1 \right) \right]=0 \ & \Leftrightarrow x\left( x-1 \right)\left( x+2 \right)=0 \ & \Leftrightarrow \left{ \begin{matrix} x=0 \ x-1=0 \ x+2=0 \ \end{matrix} \right.\Leftrightarrow \left{ \begin{matrix} x=0\left( tm \right) \ x=1\left( tm \right) \ x=-2\left( L \right) \ \end{matrix} \right. \ \end{align}](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cbegin%7Balign%7D%0A%0A%26%201%2B%5Cfrac%7B1%7D%7Bx%2B2%7D%3D%5Cfrac%7B12%7D%7B8%2B%7B%7Bx%7D%5E%7B3%7D%7D%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%201%2B%5Cfrac%7B1%7D%7Bx%2B2%7D%3D%5Cfrac%7B12%7D%7B%5Cleft(%20x%2B2%20%5Cright)%5Cleft(%204-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%20%5Cright)%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cfrac%7B%5Cleft(%20x%2B2%20%5Cright)%5Cleft(%204-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%20%5Cright)%7D%7B%5Cleft(%20x%2B2%20%5Cright)%5Cleft(%204-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%20%5Cright)%7D%2B%5Cfrac%7B4-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%7D%7B%5Cleft(%20x%2B2%20%5Cright)%5Cleft(%204-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%20%5Cright)%7D%3D%5Cfrac%7B12%7D%7B%5Cleft(%20x%2B2%20%5Cright)%5Cleft(%204-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%20%5Cright)%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%208%2B%7B%7Bx%7D%5E%7B3%7D%7D%2B4-2x%2B%7B%7Bx%7D%5E%7B2%7D%7D%3D12%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%7B%7Bx%7D%5E%7B3%7D%7D%2B%7B%7Bx%7D%5E%7B2%7D%7D-2x%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20x%5Cleft(%20%7B%7Bx%7D%5E%7B2%7D%7D-x%2B2x-2%20%5Cright)%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20x%5Cleft%5B%20x%5Cleft(%20x-1%20%5Cright)%2B2%5Cleft(%20x-1%20%5Cright)%20%5Cright%5D%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20x%5Cleft(%20x-1%20%5Cright)%5Cleft(%20x%2B2%20%5Cright)%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%3D0%20%5C%5C%0A%0Ax-1%3D0%20%5C%5C%0A%0Ax%2B2%3D0%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%3D0%5Cleft(%20tm%20%5Cright)%20%5C%5C%0A%0Ax%3D1%5Cleft(%20tm%20%5Cright)%20%5C%5C%0A%0Ax%3D-2%5Cleft(%20L%20%5Cright)%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%20%5C%5C%0A%0A%5Cend%7Balign%7D)

Vậy phương trình có nghiệm x = 0 hoặc x = 1

Điều kiện xác định: ![\left{ \begin{matrix} x-3\ne 0 \ 2x+7\ne 0 \ x+3\ne 0 \ \end{matrix} \right.\Leftrightarrow \left{ \begin{matrix} x\ne 3 \ x\ne \dfrac{-2}{7} \ x\ne -3 \ \end{matrix} \right.](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-3%5Cne%200%20%5C%5C%0A%0A2x%2B7%5Cne%200%20%5C%5C%0A%0Ax%2B3%5Cne%200%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%5Cne%203%20%5C%5C%0A%0Ax%5Cne%20%5Cdfrac%7B-2%7D%7B7%7D%20%5C%5C%0A%0Ax%5Cne%20-3%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.)

![\begin{align} & \frac{13}{\left( x-3 \right)\left( 2x+7 \right)}+\frac{1}{2x+7}=\frac{6}{\left( x-3 \right)\left( x+3 \right)} \ & \Leftrightarrow \frac{13\left( x+3 \right)}{\left( x-3 \right)\left( x+3 \right)\left( 2x+7 \right)}+\frac{\left( x-3 \right)\left( x+3 \right)}{\left( x-3 \right)\left( x+3 \right)\left( 2x+7 \right)}=\frac{6\left( 2x+7 \right)}{\left( x-3 \right)\left( x+3 \right)\left( 2x+7 \right)} \ & \Leftrightarrow 13\left( x+3 \right)+\left( x-3 \right)\left( x+3 \right)=6\left( 2x+7 \right) \ & \Leftrightarrow 13x+39+{{x}^{{2}}-9=12x+42 \
& \Leftrightarrow {{x}}{2}}+x-12=0 \ & \Leftrightarrow {{x}^{2}}+4x-3x-12=0 \ & \Leftrightarrow x\left( x+4 \right)-3\left( x+4 \right)=0 \ & \Leftrightarrow \left( x-3 \right)\left( x+4 \right)=0 \ & \Leftrightarrow \left[ \begin{matrix} x-3=0 \ x+4=0 \ \end{matrix}\Leftrightarrow \left[ \begin{matrix} x=3\left( L \right) \ x=-4\left( tm \right) \ \end{matrix} \right. \right. \ \end{align}](https://https://i0.wp.com/tex.vdoc.vn/?tex=%5Cbegin%7Balign%7D%0A%0A%26%20%5Cfrac%7B13%7D%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%202x%2B7%20%5Cright)%7D%2B%5Cfrac%7B1%7D%7B2x%2B7%7D%3D%5Cfrac%7B6%7D%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B3%20%5Cright)%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cfrac%7B13%5Cleft(%20x%2B3%20%5Cright)%7D%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B3%20%5Cright)%5Cleft(%202x%2B7%20%5Cright)%7D%2B%5Cfrac%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B3%20%5Cright)%7D%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B3%20%5Cright)%5Cleft(%202x%2B7%20%5Cright)%7D%3D%5Cfrac%7B6%5Cleft(%202x%2B7%20%5Cright)%7D%7B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B3%20%5Cright)%5Cleft(%202x%2B7%20%5Cright)%7D%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%2013%5Cleft(%20x%2B3%20%5Cright)%2B%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B3%20%5Cright)%3D6%5Cleft(%202x%2B7%20%5Cright)%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%2013x%2B39%2B%7B%7Bx%7D%5E%7B2%7D%7D-9%3D12x%2B42%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%7B%7Bx%7D%5E%7B2%7D%7D%2Bx-12%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%7B%7Bx%7D%5E%7B2%7D%7D%2B4x-3x-12%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20x%5Cleft(%20x%2B4%20%5Cright)-3%5Cleft(%20x%2B4%20%5Cright)%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cleft(%20x-3%20%5Cright)%5Cleft(%20x%2B4%20%5Cright)%3D0%20%5C%5C%0A%0A%26%20%5CLeftrightarrow%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%0Ax-3%3D0%20%5C%5C%0A%0Ax%2B4%3D0%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%5CLeftrightarrow%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%0Ax%3D3%5Cleft(%20L%20%5Cright)%20%5C%5C%0A%0Ax%3D-4%5Cleft(%20tm%20%5Cright)%20%5C%5C%0A%0A%5Cend%7Bmatrix%7D%20%5Cright.%20%5Cright.%20%5C%5C%0A%0A%5Cend%7Balign%7D)

Vậy là nghiệm duy nhất của phương trình.

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Trên đây là lời giải chi tiết bài tập Toán 8 bài 5: Phương trình chứa ẩn ở mẫu cho các em học sinh tham khảo, nắm được cách giải các dạng toán Chương 3: Phương trình bậc nhất một ẩn Toán 8 Tập 2. Với lời giải hướng dẫn chi tiết các bạn có thể so sánh kết quả của mình từ đó nắm chắc kiến thức Toán lớp 8. Chúc các bạn học tốt và nhớ thường xuyên tương tác với GiaiToan để có thêm nhiều tài liệu chất lượng miễn phí nhé!

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