Tham khảo nha:

https://hoc247.net/hoi-dap/toan-9/giai-he-phuong-trinh-y-xy-2-6x-2-1-x2y-2-5x-2-faq361806.html

các bn ơi giúp mình với

a: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\cdot\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=1+2=3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\cdot3x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\left(2-2y\right)-3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4-7y=18\\3x=2-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7y=-14\\3x=2-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\3x=2-2\cdot\left(-2\right)=6\end{matrix}\right.\)

=>x=2 và y=-2

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

Lời giải:

Nếu \(x=0\Rightarrow x^2y^2=-1\) (vô lý)

Nếu \(y=0\Rightarrow 6x^2=0\Leftrightarrow x=0\).Thay vào pt (2) thì \(1=5x^2=0\) (vô lý)

Vậy \(x,y\neq 0\)

PT tương đương: \(\left\{\begin{matrix} y(1+xy)=6x^2\\ (xy+1)^2-2xy=5x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} xy+1=\frac{6x^2}{y}\\ (xy+1)^2-2xy=5x^2\end{matrix}\right.\)

\(\Rightarrow \left(\frac{6x^2}{y}\right)^2-2xy=5x^2\)

\(\Leftrightarrow \frac{36x^3}{y^2}-2y=5x\) (do \(x\neq 0\) )

\(\Leftrightarrow 36x^3-2y^3=5xy^2\)

Đặt \(x=ty\Rightarrow 36t^3y^3-2y^3-5ty^3=0\)

\(\Leftrightarrow 36t^3-2-5t=0\) (do \(y\neq 0\) )

\(\Leftrightarrow (2t-1)(18t^2+9t+2)=0\)

Thấy rằng \(18t^2+9t+2=18(t+\frac{1}{4})^2+\frac{7}{8}>0\) nên \(2t-1=0\)

\(\Leftrightarrow t=\frac{1}{2}\Leftrightarrow x=\frac{y}{2}\Leftrightarrow 2x=y\)

Thay vào PT (1)

\(2x+4x^3=6x^2\Leftrightarrow 1+2x^2-3x=0\) (do x khác 0)

\(\Leftrightarrow (2x-1)(x-1)=0\)

Nếu \(x=\frac{1}{2}\Rightarrow y=1\)

Nếu \(x=1\Rightarrow y=2\)

Thử lại thấy thỏa mãn.

Vậy \((x,y)\in \left\{(\frac{1}{2};1); (1;2)\right\}\)

Câu hỏi của Thuý Lady - Toán lớp 9 - Học toán với OnlineMath