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b) Ta có: \(x^3-x^2y-xy^2+y^3\)
\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)^2\)
a: (3x-5)^2-(x+3)^2
=(3x-5-x-3)(3x-5+x+3)
=(2x-8)(4x-2)
=2(x-4)*2*(2x-1)
=4(x-4)(2x-1)
b: (2x+1)^2-4(x-3)^2
=(2x+1)^2-[2*(x-3)]^2
=(2x+1)^2-(2x-6)^2
=(2x+1-2x+6)(2x+1+2x-6)
=(4x-5)*7
\(Bài1:\\ a,\left(4x-1\right)\left(2x^2-x-1\right)=4x\left(2x^2-x-1\right)-\left(2x^2-x-1\right)=8x^3-4x^2-4x-2x^2+x+1=8x^3-6x^2-3x+1\\ b,\left(4x^3+8x^2-2x\right):2x\\ =2x\left(2x^2+4x-1\right):2x\\ =2x^2+4x-1\)
\(Bài2:\\ a,2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\\ b,2xy+2x+yz+z=2x\left(y+1\right)+z\left(y+1\right)=\left(y+1\right)\left(2x+z\right)\\ c,x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\)
\(a,=\left(x-1\right)^4-2\left(x-1\right)^2+1\\ =\left[\left(x-1\right)^2-1\right]^2\\ =\left(x^2-2x-2\right)^2\\ b,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-4\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36\\ =\left(x^2+6x+4\right)\left(x^2+6x+9\right)\\ =\left(x+3\right)^2\left(x^2+6x+4\right)\)
a: \(=x^2\left(x-2\right)\)
b: \(=\left(x-3\right)\left(2x-9\right)\)
\(a,=x^2\left(x-2\right)\\ b,=\left(x-3\right)\left(2x-9\right)\\ c,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
b) x-2x√x+x
\(\Leftrightarrow x\left(1-2\sqrt{x}+1\right)\)
\(\Leftrightarrow x\left(2-2\sqrt{x}\right)\)
\(\Leftrightarrow2x\left(1-\sqrt{x}\right)\)