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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 )
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Ta có: \(x^2+y^2+2xy+x+y-6\)
\(=\left(x+y\right)^2+x+y-6\)
\(=\left(x+y\right)^2+x+y-9+3\)
\(=\left[\left(x+y\right)^2-3^2\right]+\left(x+y+3\right)\)
\(=\left(x+y-3\right)\left(x+y+3\right)+\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-2\right)\)
\(=\left(\sqrt{2x}\right)^2-\left(\sqrt{y}\right)^2\)
\(=\left(\sqrt{2x}-\sqrt{y}\right)\left(\sqrt{2x}+\sqrt{y}\right)\)
3\(x\) - y
= (\(\sqrt{3x}\))2 - (\(\sqrt{y}\))2
= (\(\sqrt{3x}\) - \(\sqrt{y}\)).(\(\sqrt{3x}\) + \(\sqrt{y}\))
\(x^4+y^4\)
\(=x^4+2x^2y^2+y^4-2x^2y^2\)
\(=\left(x^2+y^2\right)^2-\left(\sqrt{2}xy\right)^2\)
\(=\left(x^2+\sqrt{2}xy+y^2\right)\left(x^2-\sqrt{2}xy+y^2\right)\)
Ta có \(4x-5\sqrt{x}-3\) = (\(4x-\frac{2×2×5\sqrt{x}}{2×2}+\frac{25}{16}\)) - \(\frac{73}{16}\)
= (\(2\sqrt{x}-\frac{5}{4}\))2 - \(\frac{73}{16}\)
= (\(2\sqrt{x}-\frac{5}{4}-\frac{\sqrt{73}}{4}\))(\(2\sqrt{x}-\frac{5}{4}+\frac{\sqrt{73}}{4}\))
Trả lời :
3( x - y ) - 5x( x - y )
= ( 3 - 5x ) . ( x - y )
3(x-y)-5x(x-y)
= ( 3 - 5x )( x - y )
Học tốt