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a) \(x\left(x-1\right)+\left(1-x\right)^2\)
\(=x\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x-1\right)\left(x+x-1\right)\)
\(=\left(x-1\right)\left(2x-1\right)\)
b) \(\left(x+1\right)^2-3\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)-3\right]\)
\(=\left(x+1\right)\left(x+1-3\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
c) \(2x\left(x-2\right)-\left(x-2\right)^2\)
\(=\left(x-2\right)\left[2x-\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(2x-x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)\)
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
a: Ta có: \(x^5-x^3+x^2-1\)
\(=x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x-1\right)\cdot\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
b: Ta có: \(5x^3-45x\)
\(=5x\left(x^2-9\right)\)
\(=5x\left(x-3\right)\left(x+3\right)\)
c: Ta có: \(16x^4y^2+2xy^5\)
\(=2xy^2\left(8x^3+y^3\right)\)
\(=2xy^2\cdot\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
d: Ta có: \(a^3-8+6a^2-12a\)
\(=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+8a+4\right)\)
e: Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
c: \(=\left(5x-y\right)\left(5x+y\right)\)
e: \(=\left(x-2\right)\left(x-3\right)\)
a) x(4y-10x)
b)3(x+2y)+(x+1)
c)(5x-y)(5x+y)
d)5x(y-z)2
e)(x-3)(x-2)
f)(2x+y)3
a) \(4x^2-1=\left(2x+1\right)\left(2x-1\right)\)
b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)
c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)
\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)
\(=3b\left(2a-b\right)\)
`a, 4x^2-1 = (2x+1)(2x-1)`
`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`
`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`
a: =(x-z)(y+8)
b; =x^2-2x-3x+6
=(x-2)(x-3)
c: =x^4+10x^2-x^2-10
=(x^2+10)(x^2-1)
=(x^2+10)(x-1)(x+1)
a: \(x^2-8x+16x=x^2+8x=x\left(x+8\right)\)
b: \(4x^2-8xyz+4y^2=4\left(x^2-2xyz+y^2\right)\)
c: \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
a, \(=\left(xy+1+x-y\right)\left(xy+1-x+y\right)\)
b, \(\left(x+y-x+y\right)[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2]\)
\(=2y[x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2]\)
\(=2y\left(3x^2+y^2\right)\)
c,\(=3\left(x+1\right)^2\left(x^2-x+1\right)y^2\)
câu a, b áp dụng hằng đẳng thức rồi làm nha
c) 3x4y2 + 3x3y2 + 3xy2 + 3y2
= ( 3x4y2 + 3x3y2 ) + ( 3xy2 + 3y2 )
= 3x3y2 ( x + 1) + 3y2 ( x + 1 )
= ( 3x3y2 + 3y2 ) ( x + 1 )
= 3y2 ( x3 + 1 ) ( x + 1 )
= 3y2 ( x + 1 ) ( x2 - x + 1 ) ( x + 1 )
= 3y2 ( x + 1 )2 ( x2 - x + 1 )
a) \(x^8+x^4-2\)
\(=x^8+x^7+x^6+x^5+2x^4+2x^3+2x^2+2x-x^7-x^6-x^5-x^4-2x^3-2x^2-2x-2\)
\(=x\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)-\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)
\(=\left(x-1\right)\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)
\(=\left(x-1\right)\left[x^4\left(x^3+x^2+x+1\right)+2\left(x^3+x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^4+2\right)\left(x^3+x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^2+2\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+1\right)\left(x^2+1\right)\left(x+1\right)\)
c) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=x^4+2x^3+x^2-2x^2-2x-15\)
\(=x^4+2x^3-x^2-2x-15\)
\(=x^4+x^3+3x^2+x^3+x^2+3x-5x^2-5x-15\)
\(=x^2\left(x^2+x+3\right)+x\left(x^2+x+3\right)-5\left(x^2+x+3\right)\)
\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)