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1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) \(4\left(x^2-y^2\right)+4x+1\)
\(=4x^2-4y^2+4x+1\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1^2\right]-\left(2y\right)^2\)
\(=\left(2x+1\right)^2-\left(2y\right)^2\)
\(=\left(2x-2y+1\right)\left(2x+2y+1\right)\)
b) \(x^2+1-x^3-x^2\)
\(=1-x^3\)
\(=\left(1-x\right)\left(1+x+x^2\right)\)
bạn hỏi từng câu 1 lần thôi cũng đc hỏi 1 lần 17 câu thì thánh nào vô kiên nhẫn trả lời hết đc ^^
\(a.=x^3-2x^2+x^2-2x+x-2=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b.\(=2x^3+x^2-2x^2-x-2x-1=x^2\left(2x+1\right)-x\left(2x-1\right)-\left(2x-1\right)\)\(=\left(2x-1\right)\left(x^2-x-1\right)\)
c.\(3x^3-x^2+6x^2-2x-12x+4=x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2+2x-4\right)\)
d.\(3x^3-x^2-6x^2+2x+15x-5=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
t i c k cho mình nha
Bài 1: 4a2-4ab+b2-9a2b2
=(2a)2-2.2a.b+b2-(3ab)2
=(2a-b)2-(3ab)2
=(2a-b-3ab)(2a-b+3ab)
a/ (4a2-4ab+b2)-9a2b2
= (2a-b)2-(3ab)2
= (2a-b-3ab) (2a-b+3ab)
a) \(4x^4 + 4x^3 + 5x^2 + 2x + 1\)
\(= 4x^4 + 4x^3 + x^2 + 4x^2 + 2x + 1\)
\(= (2x^2 + x)^2 + (2x + 1)^2\)
\(= x(2x + 1)^2 + (2x + 1)^2\)
\(= (x + 1)(2x + 1)^2\)
a) 4x4 + 4x3 + 5x2 + 2x + 1 = 4x4 + 4x3 + x2 + 4x2 + 2x + 1
= (4x4 + 4x3 + x2) + ( 4x2 + 2x + 1)
= x2 (2x2 + 4x + 1) + (2x)2 + 2x +1
= x2 (2x + 1)2 + 2x ( 2x + 1) + 1
= \([\)(2x + 1)x\(]\)2 + 2 (2x+1) x + 12
= \([\)(2x + 1)x + 1\(]\)2