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\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Ta có x + y + z = 0
<=> (x + y + z)2 = 0
<=> \(x^2+y^2+z^2+2xy+2yz+2zx=0\)
\(\Leftrightarrow xy+yz+zx=-3\) (vì x2 + y2 + z2 = 6)
\(\Leftrightarrow x\left(y+z\right)+yz=-3\)
\(\Leftrightarrow-x^2+yz=-3\Leftrightarrow yz=x^2-3\) (vì x + y + z = 0)
Khi đó \(x^3+y^3+z^3=x^3+(y+z).(y^2+z^2-yz)\)
\(=x^3-x.[6-x^2-(x^2-3)]\)
\(=x^3-x.(9-2x^2)=3x^3-9x=6\)
Ta được \(\Leftrightarrow x^3-3x-2=0\Leftrightarrow(x^3+1)-3(x+1)=0\)
\(\Leftrightarrow(x+1)(x^2-x-2)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Với x = -1 ta có hệ \(\left\{{}\begin{matrix}y+z=1\\y^2+z^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\(1-z)^2+z^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\z^2-z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\\left[{}\begin{matrix}z=-1\\z=2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\z=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\z=2\end{matrix}\right.\end{matrix}\right.\)
Với x = 2 ta có hệ : \(\left\{{}\begin{matrix}y+z=-2\\y^2+z^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\(-2-z)^2+z^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\z^2+2z+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\z=-1\end{matrix}\right.\Leftrightarrow y=z=-1\)
Vậy (x;y;z) = (2;-1;-1) ; (-1 ; 2 ; -1) ; (-1 ; -1 ; 2)
a/ \(\frac{3x^2-11x+8}{2x^2-9x+7}=\frac{\left(x-1\right)\left(3x-8\right)}{\left(x-1\right)\left(2x-7\right)}=\frac{3x-8}{2x-7}\)
câu b,c tương tự nha ^^
