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\(4\left(1+x\right)\left(1+y\right)\left(1+x+y\right)-3x^2y^2=4\left(1+x+y+xy\right)\left(1+x+y\right)-3x^2y^2\)
\(=4\left(1+x+y\right)^2+4xy\left(1+x+y\right)+x^2y^2-4x^2y^2\)
\(=\left[2\left(1+x+y\right)+xy\right]^2-\left(2xy\right)^2=\left(2+2x+2y+xy-2xy\right)\left(2+2x+2y+xy+2xy\right)\)
\(=\left(2+2x+2y-xy\right)\left(2+2x+2y+3xy\right)\)
giúp mình câu khác được ko? câu này mình biết làm òi
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a) x4 + 2x3 + x2
= x2 ( x2 + 2x + 1 )
= x2 ( x + 1 )2
b) 5x2 - 10xy + 5y2 - 20z2
= 5 [(x2 - 2xy + y2 ) - 4z2 ]
= 5 [( x - y )2 - ( 2z )2 ]
= 5 ( x - y - 2z ) ( x - y + 2z )
c) x3 - x + 3x2y + 3xy2+ y3- y
= ( x3 + 3x2y + 3xy2 + y3 ) - ( x + y )
= (x + y )3 - ( x + y)
= ( x + y ) [( x + y )2 - 1 ]
= ( x + y ) ( x + y + 1 ) ( x + y - 1 )
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\(a,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ b,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
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a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
b: \(B=5x^2-7x\sqrt{y}+2y\)
\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)
\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)
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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:
\(x^2\left(x^2-4y+3\right)+\left(x^2-2xy+x\right)^2=0\)
\(\Leftrightarrow x^2\left(x^2-4y+3\right)+x^2\left(x-2y+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x^2-4y+3+\left(x-2y+1\right)^2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)
\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)
\(\Leftrightarrow-2y+4y^2-8y+4=0\)
\(\Leftrightarrow...\)
\(x^4-2y^4-x^2y^2+x^2+y^2=\left(x^4-y^4\right)-\left(x^2y^2-x^2\right)+\left(y^2-y^4\right)=\left(x^2-y^2\right)\left(x^2+y^2\right)-x^2\left(y^2-1\right)-y^2\left(y^2-1\right)=\left(x^2+y^2\right)\left(x^2-y^2\right)-\left(y^2-1\right)\left(x^2+y^2\right)=\left(x^2+y^2\right)\left(x^2-y^2-y^2+1\right)=\left(x^2+y^2\right)\left(x^2-2y^2+1\right)\)