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a) \(16x^2-\left(x^2+4\right)^2=\left(4x\right)^2-\left(x^2+4\right)^2\)
\(=\left(4x+x^2+4\right)\left(4x-x^2-4\right)\)
\(=\left(x+2\right)^2\left\{-\left(x^2-4x+4\right)\right\}=\left(x+2\right)^2\left\{-\left(x-2\right)^2\right\}\)
Ở đây mình không đổi \(-\left(x-2\right)^2=\left(2-x\right)^2\)được vì vốn dĩ \(\left(x-2\right)^2=\left(2-x\right)^2\)
b) \(\left(x^2+9\right)^2-36=\left(x^2+9\right)^2-6^2\)
\(=\left(x^2+9+6\right)\left(x^2+9-6\right)=\left(x^2+15\right)\left(x^2-3\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
f) \(x^4y^4-z^4=\left\{\left(x^2y^2\right)^2-\left(z^2\right)^2\right\}\)
\(=\left\{\left(xy\right)^2-z^2\right\}\left\{x^2y^2+z^2\right\}\)
\(=\left(xy-z\right)\left(xy+z\right)\left(x^2y^2+x^2\right)\)
\(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x-y\right)^2-4\left(4x-y\right)\)
\(=\left(4x-y\right)\left(4x-y-4\right)\)
a) \(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x\right)^2-8xy+y^2+4\left(y-4x\right)\)
\(=\left(4x-y\right)^2+4\left(y-4x\right)\)
\(=\left(y-4x\right)^2+4\left(y-4x\right)=\left(y-4x\right)\left(y-4x+4\right)\)
\(a,6x^3-9x^2=3x^2\left(2x-3\right)\)
\(b,4x^2y-8xy^2+10x^2y^2=2xy\left(2x-4y+5xy\right)\)
\(c,20x^2y-12x^3=4x^2\left(5y-3x\right)\)
\(d,4xy^2+8xyz=4xy\left(y+2z\right)\)
\(-8x^2y^2-12xy^3-4xy^2\)
\(=-8x^2y^2-8xy^3-4xy^3-4xy\)
\(=-8xy\left(xy-y^2\right)-4xy\left(y^2-1\right)\)
\(=-8xy\left(y\left(x-y\right)\right)-4xy\left(y-1\right)\left(y+1\right)\)
\(=-4.2xy\left(y\left(x-y\right)\right)-4xy\left(y-1\right)\left(y+1\right)\)
\(=-4\left(2xy\left(y\left(x-y\right)\right)-xy\left(y-1\right)\left(y+1\right)\right)\)
Vậy thôi thành nhân tử là dc rồi
Ủng hộ nha
Thanks
\(12xy-4x^2y+8xy^2\)
\(=4xy\left(3-x+2y\right)\)
\(-125a^3+75a^2-15a+1\)
\(=-\left(125a^3-75a^2+15a-1\right)\)
\(=-\left[\left(5a\right)^3-3.\left(5a\right)^2.1+3.5a.1^3-1^3\right]\)
\(=-\left(5a-1\right)^3\)