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

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

\(\Rightarrow9x^2+y^2-6xy+6x-2y+1=9\)

\(\Leftrightarrow\left(3x-y+1\right)^2=9\)

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

Đến đây chia 2 trường hợp và thế vào 1 trong 2 pt để giải

- Trừ hai pt ta được :\(x^3-y^3-x^2+y^2+x-y+1-1=2y-2x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x^2+xy+y^2-x-y+3=0\end{matrix}\right.\)

TH1 : x = y

PT ( I ) TT : \(x^3-x^2+x+1-2x=x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow x=y=\pm1\)

TH₂ : \(x^2+xy+y^2-x-y+3=0\)

\(\Leftrightarrow x^2+\dfrac{y^2}{4}+\dfrac{1}{4}+xy-x-\dfrac{1}{2}y+\dfrac{3}{4}y^2-\dfrac{1}{2}y+\dfrac{11}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2+\dfrac{8}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2=-\dfrac{8}{3}\left(VL\right)\)

Vậy ....

\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{2}\\\dfrac{x+5}{2}=\dfrac{x+7}{3}-4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\left(2x+1\right)}{12}-\dfrac{4\left(y-2\right)}{12}=\dfrac{6}{12}\\\dfrac{3\left(x+5\right)}{6}=\dfrac{2\left(x+7\right)}{6}-\dfrac{24}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3\left(2x+1\right)-4\left(y-2\right)=6\\3\left(x+5\right)=2\left(x+7\right)-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+3-4y+8=6\\3x+15=2y+14-24\end{matrix}\right.\\ \Leftrightarrow\Leftrightarrow\left\{{}\begin{matrix}6x-4y+11=6\\3x+15=2y-10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-4y=-5\\3x-2y=-25\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2\left(3x-2y\right)=-5\\3x-2y=-25\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x-2y=-\dfrac{5}{2}\\3x-2y=-25\left(vô.lí\right)\end{matrix}\right.\)

Vậy hệ phương trình vô nghiệm

Mấy hệ pt của bạn đọc không ra bạn ơi. B ghi lại đi nhấp vô chỗ \(\sum\) để ghi công thức nhé

\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{1}{4}-\dfrac{y}{3}+\dfrac{2}{3}=\dfrac{1}{12}\\\dfrac{x}{2}+\dfrac{5}{2}-\dfrac{y}{3}-\dfrac{7}{3}=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{5}{6}\\\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{25}{6}\end{matrix}\right.\) (vô lý)

Vậy HPT vô nghiệm

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

Đặt \(\dfrac{1}{2x+y}=a;\sqrt{y}=b\)

ĐK: x, y ≥ 0

⇒ \(\left\{{}\begin{matrix}a+b=2\\3a+2b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

Trả ẩn: \(\left\{{}\begin{matrix}\dfrac{1}{2x+y}=1\\\sqrt{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x+1}=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy hpt có nghiệm (x ; y) = (0 ; 1)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2x-4+2}{x-2}+\dfrac{y+1+1}{y+1}=\dfrac{26}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=\dfrac{26}{5}-1-2=\dfrac{11}{5}\end{matrix}\right.\)

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