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\(\left(x+1\right)\left(y+1\right)=8\\ \Rightarrow xy+x+y+1=8\\ \Rightarrow xy+x+y=7\)
\(x\left(x+1\right)+y\left(y+1\right)+xy=17\\ \Rightarrow x^2+y^2+x+y+xy=17\\ \Rightarrow x^2+y^2=10\)
a) \(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)+\left(-30x^2+30x-30\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x-30\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+y+z=2\left(1\right)\\2xy-z^2=4 \left(2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2xy+2yz+2xz=4\\2xy-z^2=4\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=2xy-z^2\)
\(\Leftrightarrow x^2+y^2+2z^2+2yz+2xz=0\)
\(\Leftrightarrow\left(x+z\right)^2+\left(y+z\right)^2=0\)
\(\Rightarrow x=y=-z\) thay vào (1) ta được : \(-z-z+z=2\Rightarrow z=-2\)
\(\Rightarrow x=y=2\)
Vậy \(x=y=2;z=-2\)
a)\(\hept{\begin{cases}2x-3y=1\\4x-5y=2\end{cases}\Leftrightarrow\hept{\begin{cases}4x-6y=2\\4x-5y=2\end{cases}}}\)
Trừ 2 vế lại ta được
\(4x-4x-6y+5y=0\Leftrightarrow-y=0\Leftrightarrow y=0\)
\(\Rightarrow x=\frac{1}{2}\)
`{((a-1)x+y=a),(x+(a-1)y=2):}`
`<=>{(ax-x+y=a),(x+ay-y=2):}`
`<=>{(a(x-1)=x-y<=>a=[x-y]/[x-1]),(x+[x-y]/[x-1]-y=2):}`
`<=>x(x-1)+x-y-y(x-1)=2(x-1)`
`<=>x^2-x+x-y-xy+y=2x-2`
`<=>x^2-xy-2x+2=0`
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`b)x^2-xy-2x+2=0`
`<=>xy=x^2-2x+2`
`<=>y=x-2+2/x`
Thay `y=x-2+2/x` vào `6x^2-17y=7` có:
`6x^2-17(x-2+2/x)=7`
`<=>6x^3-17x^2+34x-34-7x=0`
`<=>6x^3-12x^2-5x^2+10x+17x-34=0`
`<=>(x-2)(6x^2-5x+17)=0`
Mà `6x^2-5x+17 > 0`
`=>x-2=0<=>x=2`
`=>y=2-2+2/2=1`
Thay `x=2;y=1` vào `(a-1)x+y=a` có: `(a-1).2+1=a<=>a=1`
\(\left\{{}\begin{matrix}x^2-4y^2=24\\\left(5-2y\right)\left(x-7\right)=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=7\\4y^2=49-24=25=>\left|y\right|=\dfrac{5}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x^2-25=24=>x^2=49=>\left|x\right|=7\end{matrix}\right.\)