a.

Thay số 12 từ pt trên xuống dưới:

\(x^3+2xy^2+y\left(x^2+8y^2\right)=0\)

\(\Leftrightarrow x^3+x^2y+2xy^2+8y^3=0\)

\(\Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\)

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

Thế vào pt đầu:

\(\left(-2y\right)^2+8y^2=12\Leftrightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)

b.

Thế số 1 từ pt trên xuống dưới:

\(x^7+y^7=\left(x^4+y^4\right)\left(x^3+y^3\right)\)

\(\Leftrightarrow x^4y^3+x^3y^4=0\)

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

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

Thế vào pt đầu: \(\Rightarrow\left[{}\begin{matrix}y^3=1\\x^3=1\\x^3-x^3=1\left(vô-nghiệm\right)\end{matrix}\right.\)

Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-y=0\\2x-y+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x^2\\y=2x+1\end{matrix}\right.\)

Thế vào pt đầu:

\(\left[{}\begin{matrix}x^3+x-2=0\\x\left(2x+1\right)+x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x^2+x+2\right)=0\\x^2+x-1=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

\(x^2-2xy+x=-y\)

Thế vào \(y^2\) ở pt dưới:

\(x^2\left(x^2-4y+3\right)+\left(x^2-2xy+x\right)^2=0\)

\(\Leftrightarrow x^2\left(x^2-4y+3\right)+x^2\left(x-2y+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x^2-4y+3+\left(x-2y+1\right)^2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)

\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)

\(\Leftrightarrow-2y+4y^2-8y+4=0\)

\(\Leftrightarrow...\)

Bạn tham khảo lời giải tại đây:

Câu hỏi của Nguyễn Thu Ngà - Toán lớp 9 | Học trực tuyến

\(1.\left(x\ne\pm1\right)\Rightarrow pt\Leftrightarrow\left(x-m\right)\left(x-1\right)=\left(x+1\right)\left(x-2\right)\)

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

\(\Leftrightarrow-x\left(m+1\right)+m=-x-2\)

\(\Leftrightarrow x=\dfrac{m+2}{m}\left(m\ne0\right)\)

\(pt-có-ngo-duy-nhất\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m+2}{m}\ne1\\\dfrac{m+2}{m}\ne-1\end{matrix}\right.\)\(\Leftrightarrow m\ne-1\)

\(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m\ne-1\end{matrix}\right.\)

\(2.\left\{{}\begin{matrix}x^2+8y^2=12\left(1\right)\\x^3+2xy^2+12y=0\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow x^3+2xy^2+y\left(x^2+8y^2\right)=0\)

\(\Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2y\left(3\right)\\x^2-xy+4y^2=\left(x-\dfrac{y}{2}\right)^2+\dfrac{15}{4}y^2=0\left(4\right)\end{matrix}\right.\)

\(\left(3\right)\left(1\right)\Rightarrow4y^2+8y^2=12\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)

với \(x=y=0\) không là nghiệm của hệ pt

với \(x=y\ne0\Rightarrow\left(4\right)>0\Rightarrow\left(4\right)-vô-nghiệm\)

\(\Rightarrow\left(x;y\right)=\left\{\left(-2;1\right);\left(2;-1\right)\right\}\)

\(1,\Leftrightarrow\left(x-m\right)\left(x-1\right)=x^2-x-2\\ \Leftrightarrow x^2-x-mx+m-x^2+x+2=0\\ \Leftrightarrow mx=m+2\)

PT có nghiệm duy nhất \(\Leftrightarrow m\ne0\)

\(2,\Leftrightarrow\left\{{}\begin{matrix}x^2y+8y^3=12y\\x^3+2xy^2+12y=0\end{matrix}\right.\)

Thế \(PT\left(1\right)\rightarrow PT\left(2\right)\Leftrightarrow x^3+2xy^2+x^2y+8y^3=0\)

\(\Leftrightarrow\left(x+2y\right)\left(x^2-2xy+4y^2\right)+xy\left(x+2y\right)=0\\ \Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2y\\\left(x-\dfrac{1}{2}y\right)^2+\dfrac{15}{4}y^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2y\\\left\{{}\begin{matrix}x-\dfrac{1}{2}y=0\\y^2=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2y\\x=y=0\end{matrix}\right.\)

Thay \(x=y=0\Leftrightarrow0+0=12\left(loại\right)\)

Thay \(x=-2y\Leftrightarrow4y^2+8y^2=12y^2=12\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)

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

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

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

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

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

\(\Rightarrow y=2x\)

Thế xuống dưới:

\(x^4-2x^3-x^2+2x+1=0\)

Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)

\(x^2+\frac{1}{x^2}-2\left(x-\frac{1}{x}\right)-1=0\)

Đặt \(x-\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2+2\) pt trở thành:

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

\(\Leftrightarrow x-\frac{1}{x}=1\Leftrightarrow x^2-x-1=0\Leftrightarrow...\)

1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)

=> Hệ có vô số nghiệm.

3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)