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Ta có: \(x^3-\left(y-2\right)^3+\left(y-x-2\right)^2\)
\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4\right)+\left(x-y+2\right)^2\)
\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4+x-y+2\right)\)
\(=\left(x-y+2\right)\left(x^2+y^2+6+xy-x-5y\right)\)
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: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử
= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung
= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung
c: \(x^2-4+3\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(3x-6\right)\)
\(=\left(x-2\right)\left(x+2+3x-6\right)\)
\(=\left(4x-4\right)\left(x-2\right)\)
\(=4\left(x-1\right)\left(x-2\right)\)
\(x^3-y^3+2x^2+2xy\)
\(=x\left(x^2-y^2+2x+2y\right)\)
\(=\)\(x\left[\left(x+y\right)\left(x-y\right)+2\left(x+y\right)\right]\)
\(=x\left(x+y\right)\left(x-y+2\right)\)
`a. =4(x^2+4x+3)=4(x^2+3x+x+3)=4(x+3)(x+1)`
`b. =x^2+8x-7x-56=x(x+8)-7(x+8)=(x+8)(x-7)`
`c. =x^2-9x+8x-72=x(x-9)+8(x-9)=(x-9)(x+8)`
`d. =(x-y)^2-9=(x-y-3)(x-y+3)`
a: =x(4x^2+4x+1)
=x(2x+1)^2
b: =(x-y)^2-49
=(x-y-7)(x-y+7)
a)\(=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x-y+1\right)\left(x+y+1\right)\)
b)\(=16-\left(x^2-2xy+y^2\right)=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)
\(x^3-4x-12+3x^2=x\left(x^2-2^2\right)+3\left(x^2-2^2\right)=\left(x-2\right)\left(x+2\right)\left(x+3\right)\)
\(x^2+2xy-15y^2=x^2+2xy+y^2-16y^2=\left(x+y\right)^2-\left(4y\right)^2=\left(x-3y\right)\left(x+5y\right)\)
\(\left(x-y\right)^2-6\left(x-y\right)-16=\left(x-y\right)^2-2\times\left(x-y\right)\times3+9-25=\left(x-y-3\right)^2-5^2=\left(x-y-8\right)\left(x-y+2\right)\)