NA

\(\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}-\dfrac{3y}{x+1}=5\\\dfrac{4x-6}{y+2}+\dfrac{5y}{x+1}=3\end{matrix}\right.\)

H24
17 tháng 9 2022 lúc 22:46

`{([2x-3]/[y+2]-[3y]/[x+1]=5),([4x-6]/[y+2]+[5y]/[x+1]=3):}`  `ĐK: x \ne -1,y \ne -2`

`<=>{([2x-3]/[y+2]-3[y]/[x+1]=5),(2[2x-3]/[y+2]+5[y]/[x+1]=3):}`

Đặt `[2x-3]/[y+2]=a;y/[x+1]b` khi đó hệ ptr có dạng:

    `{(a-3b=5),(2a+5b=3):}`

`<=>{(a=34/11),(b=-7/11):}`

  `=>{([2x-3]/[y+2]=34/11),(y/[x+1]=-7/11):}`

`<=>{(22x-33=34y+68),(11y=-7x-7):}`

`<=>{(22x-34y=101),(7x+11y=-7):}`

`<=>{(x=291/160),(y=-287/160):}`   (t/m)

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NL
17 tháng 9 2022 lúc 22:50

\(\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}-\dfrac{3y}{x+1}=5\\\dfrac{4x-6}{y+2}+\dfrac{5y}{x+1}=3\end{matrix}\right.\)

DKXD: \(x\ne1,y\ne-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}-\dfrac{3y}{x+1}=5\\\dfrac{2\left(2x-3\right)}{y+2}+\dfrac{5y}{x+1}=3\end{matrix}\right.\)

Đặt \(\dfrac{2x-3}{y+2}=a,\dfrac{y}{x+1}=b\)

Ta có hệ pt: \(\left\{{}\begin{matrix}a-3b=5\\2a+5b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{34}{11}\\b=-\dfrac{7}{11}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}=\dfrac{34}{11}\\\dfrac{y}{x+1}=-\dfrac{7}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11\left(2x-3\right)=34\left(y+2\right)\\-7\left(x+1\right)=11y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22x-33=34y+68\\-7x-7=11y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22x-34y=101\\-7x-11y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{97}{42}\\y=-\dfrac{31}{21}\end{matrix}\right.\) (tmdk)

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