Question

1.tìm m để pt có nghiệm duy nhất \(\dfrac{x-m}{x+1}=\dfrac{x-2}{x-1}\)

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

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

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