Xem lại giúp tớ dấu căn ở câu c và d nhé.  

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

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

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

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(3x+y\right)=8\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)

Chia vế cho vế: \(\frac{x^2+xy+2}{3x+y}=1\Leftrightarrow x^2+xy+2=3x+y\)

\(\Leftrightarrow x^2+\left(y-3\right)x-y+2=0\)

\(\Delta=\left(y-3\right)^2-4\left(-y+2\right)=\left(y+1\right)^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{3-y+y+1}{2}=1\\x=\frac{3-y-y-1}{2}=-y+1\end{matrix}\right.\)

- Với \(x=1\Rightarrow y^2+4y-5=0\Rightarrow y=...\)

- Với \(x=-y+1\Rightarrow3\left(-y+1\right)^2+y^2+4y\left(-y+1\right)-8=0\Rightarrow y=...\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+x=u\\y^2+y=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=8\\uv=12\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;2\right);\left(2;6\right)\)

TH1: \(\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\) \(\Rightarrow...\)

TH2: ... tương tự

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

các bn ơi giúp mình với

a) Ta có: \(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2-3y+6=5\\-4x+8+5y-15=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\2x-3y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-3\cdot0=-3\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)

Vậy: hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-24-3y-3=-2\\3x+6-2+2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y=75\\24x+16y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-25y=67\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-67}{25}\\3x=1-2y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=1-2\cdot\dfrac{-67}{25}=\dfrac{159}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\x=\dfrac{3y-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-\dfrac{3}{2};0\right)\)

b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}16x-6y=50\\9x+6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25x=53\\y=\dfrac{1-3x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\)