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Hửm đề sai rồi phải là:
`a^2+2b^2-3ab`
`=a^2-ab-2ab+2b^2`
`=a(a-b)-2b(a-b)`
`=(a-b)(a-2b)`
`a^{3}+3a^{2}-6a-8`
`=a^{3}-8+3a(a-2)`
`=(a-2)(a^{2}+2a+4)+3a(a-2)`
`=(a-2)(a^{2}+2a+4+3a)`
`=(a-2)(a^{2}+5a+4)`
`=(a-2)(a+1)(a+4)`
\(a^3-8+3a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+2a+4\right)+3a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+2a+4\right)+3a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+2a+4+3a\right)\)
\(=\left(a-2\right)\left(a^2+5a+4\right)\)
\(\left(a-2\right)\left(a+1\right)\left(a+4\right)\)
\(a,15a^2b^3+5a^3b^2=5a^2b^2\left(3b+a\right)\\ b,x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
a) 15a2b3+5a3b2=5a2b2(3b+a)
b) x2-2x+x-y2=( x2-y2)-(2x+x)
=(x-y)(x+y)-x(2-1)
=(x-y)(x+y)-x3
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
a: \(=x^2\left(x-2\right)\)
b: \(=\left(x-3\right)\left(2x-9\right)\)
\(a,=x^2\left(x-2\right)\\ b,=\left(x-3\right)\left(2x-9\right)\\ c,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
\(a^3-b^3+3a^2+3ab+b^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+3\left(a^2+ab+b^2\right)\)
\(=\left(a-b+3\right)\left(a^2+ab+b^2\right)\)
\(a^3+3a^2b+3ab^2+b^3-8c^3\)
\(=\left(a+b\right)^3-8c^3\)
\(=\left(a+b-2c\right)\left[\left(a+b\right)^2+2\left(a+b\right)c+4c^2\right]\)
\(=\left(a+b-2c\right)\left(a^2+b^2+2ab+2ac+2bc+4c^2\right)\)
thank bạn