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a)Ta có: \(x^2-6x+9=\left(x-3\right)^2\)
b) Ta có: \(27+27x+9x^2+9x^3=27\left(1+x\right)+9x^2\left(1+x\right)=\left(1+x\right)\cdot\left(27+9x^2\right)=9\left(1+x\right)\left(3+x^2\right)\)
c) Ta có: \(x^2-25-2xy+y^2=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
d) Ta có: \(7y^4-14y^3+7y^2=7y^2\left(y^2-2y+1\right)=7y^2\left(y-1\right)^2\)
e) Ta có: \(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)
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Câu 1 : x2-y2+2yz-z2=-(y2-2yz+z2-x2) Câu 2: x2-2xy+y2-xz+yz=(x2-2xy+y2)-xz+yz
=-(y-z)2 -x2 =(x-y)2-z(x-y)
=-(y-z-x)(y-z+x) =(x-y)(x-y-z)
TL
27 - x3
HT
27 - x3
= 33 - x3
Vậy x = 3