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Bài 1:
a)3x2 - 3y2 - 12x +12y=3(x2-y2)-12(x-y)=3(x-y)(x+y)-12(x-y)=3(x-y)(x+y-4)
b) 4x3 + 4xy2 + 8x2y - 16x=4x(x-4)+4xy(y+2x)=4x(x-4+y2+2xy)
c) x4 - 5x2 + 4=x4-x2-4x2+4=x2(x2-1)-4(x2-1)=(x2-1)(x2-4)=(x-1)(x+1)(x-2)(x+2)
d) x3 - 2x2 + 6x - 5=x3-x2-(x2-6x+5)=x2(x-1)-(x-1)(x-5)=(x-1)(x2-x+5)
e) x2 - 4x +3=x2-x-3x+3=x(x-1)-3(x-1)=(x-1)(x-3)
f ) 2x2 + 3x - 5=2x2-2+3x-3=2(x2-1)+3(x-1)=2(x-1)(x+1)+3(x-1)=(x-1)(2x+1)
1: A=x(x^2-1)=x(x-1)(x+1)
2: A/B=x(x+1)
3: Để A=0 thì x(x-1)(x+1)=0
hay \(x\in\left\{0;1;-1\right\}\)
4: Vì x-1;x;x+1 là ba số liên tiếp
nên x(x-1)(x+1) chia hết cho 3!
=>A chia hết cho 6
a) \(3x^2-3y^2-12x+12y\)
\(=\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-3y-12\right)\)
\(=\left(x-y\right).3.\left(x-y-4\right)\)
b) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
c) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=\left(x^4-x^2\right)-\left(4x^2-4\right)\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
Bài 1:
a) x^3 + 2x^2 + x = x.(x^2+2x+1) = x.(x+1)^2
b) xy + y^2 - x - y
= y.(x+y) - (x+y)
= (x+y).(y-1)
a)15x2+20xy2-25xy
= 5x(3x+4y-5)
b) (x+y)2- 25
= (x+y)2- 52
= (x+y+5)(x+y-5)
b) 4x2+8xy+3x-6y
= 4x(x+2y)-3(x+2y)
= (x+2y)(4x-3)
g) 3x2-6xy+3y2
e) 27+27x+9x2+x3
h) 1-4x2
= (1+4x)(1-4x)
c) 1-2y+y2
= y2+y+y-1
= y(y-1)+(y-1)
= (y-1)(y+1)
f) 8-27x3
= 23 - 3x3
= (2-3x)(22+2.3x-3x2)
= (2-3x)(4+6x-9x)
i) x6-x4+2x3-2x
= x(x5-x3+2x2-2)
= x[x3(x2-1)+2(x2-1)]
= x(x2-1)(x3+2)
Bài 5:
a: ĐKXĐ: x<>2; x<>-2
b: \(A=\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)
c: Để A=2 thì 2x-4=x+2
=>x=6