Bài 1:
a) Ta có: \(x^4-6x^3+9x^2\)
\(=x^2\left(x^2-6x+9\right)\)
\(=x^2\cdot\left(x-3\right)^2\)
b) Ta có: \(x^3-1-\left(1-x\right)\left(x-5\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(x-5\right)\)
\(=\left(x-1\right)\left(x^2+x+1+x-5\right)\)
\(=\left(x-1\right)\left(x^2+2x-4\right)\)
c) Ta có: \(36m^2-x^2+2xy-y^2\)
\(=36m^2-\left(x^2-2xy+y^2\right)\)
\(=\left(6m\right)^2-\left(x-y\right)^2\)
\(=\left(6m-x+y\right)\left(6m+x-y\right)\)
d) Ta có: \(4a^2-9b^2+2xa-3bx\)
\(=\left(2a-3b\right)\left(2a+3b\right)+x\left(2a-3b\right)\)
\(=\left(2a-3b\right)\left(2a+3b+x\right)\)
a) Ta có: x^4-6x^3+9x^2x4−6x3+9x2
=x^2\left(x^2-6x+9\right)=x2(x2−6x+9)
=x^2\cdot\left(x-3\right)^2=x2⋅(x−3)2
b) Ta có: x^3-1-\left(1-x\right)\left(x-5\right)x3−1−(1−x)(x−5)
=\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(x-5\right)=(x−1)(x2+x+1)+(x−1)(x−5)
=\left(x-1\right)\left(x^2+x+1+x-5\right)=(x−1)(x2+x+1+x−5)
=\left(x-1\right)\left(x^2+2x-4\right)=(x−1)(x2+2x−4)
c) Ta có: 36m^2-x^2+2xy-y^236m2−x2+2xy−y2
=36m^2-\left(x^2-2xy+y^2\right)=36m2−(x2−2xy+y2)
=\left(6m\right)^2-\left(x-y\right)^2=(6m)2−(x−y)2
=\left(6m-x+y\right)\left(6m+x-y\right)=(6m−x+y)(6m+x−y)
d) Ta có: 4a^2-9b^2+2xa-3bx4a2−9b2+2xa−3bx
=\left(2a-3b\right)\left(2a+3b\right)+x\left(2a-3b\right)=(2a−3b)(2a+3b)+x(2a−3b)
=\left(2a-3b\right)\left(2a+3b+x\right)=(2a−3b)(2a+3b+x)
chúc bạn thi tốt
