NT
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NT
1
9 tháng 11 2018
\(16\left(2x+3\right)^2-9\left(5x-2\right)^2\)
\(=\left[4\left(2x+3\right)\right]^2-\left[3\left(5x-2\right)\right]^2\)
\(=\left[4\left(2x+3\right)-3\left(5x-2\right)\right]\left[4\left(2x+3\right)+3\left(5x-2\right)\right]\)
\(=\left(-7x+18\right)\left(23x+6\right)\)
TT
1
20 tháng 5 2018
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)
PV
3
KN
20 tháng 6 2019
\(x^3-2x^2-5x+6\)
\(=\left(x^3-4x^2+3x\right)+\left(2x^2-8x+6\right)\)
\(=x\left(x^2-4x+3\right)+2\left(x^2-4x+3\right)\)
\(=\left(x+2\right)\left(x^2-4x+3\right)\)
55
20 tháng 6 2019
1. \(x^3-2x-5x+6\)
\(\Leftrightarrow x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+x-2\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x-1\right)\)
2. \(x^3-7x^2+15x-9\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x^2-6x\right)+\left(9x-9\right)\)
\(\Leftrightarrow x^2\left(x-1\right)-6x\left(x-1\right)+9\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-6x+9\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x\left(x-3\right)-3\left(x-3\right)\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)^2\)
HN
3
11 tháng 9 2018
1, <=> \(\left(4x\right)^2-\left(9y\right)^2\)=\(\left(4x-9y\right)\left(4x+9y\right)\)
T
11 tháng 9 2018
1) \(16x^2-81.y^2=\left(4x\right)^2-\left(9.y\right)^2=\left(4x-9y\right)\left(4x+9y\right)\)
2) \(\left(5x-3y\right)^2-\left(3x-5y\right)^2=\left(5x-3y-3x+5y\right)\left(5x-3y+3x-5y\right)=\left(2x+2y\right).\left(8x-8y\right)\)
\(=16.\left(x+y\right)\left(x-y\right)\)
3)\(4x^2-y^2+4y-4=4x^2-\left(y^2-4y+4\right)=\left(2x\right)^2-\left(y-2\right)^2=\left(2x-y+2\right).\left(2x+y-2\right)\)
4)\(9.\left(x-y\right)^2-16.\left(2x+y\right)^2=3^2.\left(x-y\right)^2-4^2.\left(2x+y\right)^2=\left(3x-3y\right)^2-\left(8x+4y\right)^2\)
\(=\left(3x-3y-8x-4y\right)\left(3x-3y+8x+4y\right)=\left(-5x-7y\right).\left(11x+y\right)\)
C
1
B
1 tháng 11 2020
a) \(9x^2-12x+4\)
\(=9x^2-6x-6x+4\)
\(=3x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(3x-2\right)^2\)
b) \(2xy+16-x^2-y^2\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y\right)^2+16\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
c) \(3x+2x^2-2\)
\(=2x^2+4x-x-2\)
\(=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
1 tháng 11 2021
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
VT
1
14 tháng 8 2019
\(2x^2\left(x-1\right)+3x^2-3x-2x+2.\)
\(2x^2\left(x-1\right)+3x\left(x-1\right)-2\left(x-1\right)\)
\(\left(x-1\right)\left(2x^2+3x-2\right)\)
\(2\left(x-1\right)\left(x^2+\frac{3}{2}x-2\right)=2\left(x-1\right)\left\{\left(x^2+\frac{2x.3}{4}+\frac{9}{16}\right)-\left(2+\frac{9}{16}\right)\right\}\)
\(2\left(x-1\right)\left\{\left(x+\frac{3}{4}\right)^2-\left(2+\frac{9}{16}\right)\right\}=2\left(x-1\right)\left\{\left(x+\frac{3}{4}-2-\frac{9}{16}\right)\left(x+\frac{3}{4}+2+\frac{9}{16}\right)\right\}\)
14 tháng 8 2019
\(=2x^3+4x^2-3x^2-6x+x+2\)
= \(2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\)
= \(\left(x+2\right)\left(2x^2-3x+1\right)\)
= \(\left(x+2\right)\left(2x^2-x-2x+1\right)\)
= \(\left(x+2\right)\left(2x\left(x-1\right)-\left(x-1\right)\right)\)
= \(\left(x+2\right)\left(x-1\right)\left(2x-1\right)\)
\(16\cdot\left(2x+3\right)^2-9\cdot\left(5x-2\right)^2\\ =\left(8x+12\right)^2-\left(15x-6\right)^2\\ =\left(8x+12-15x+6\right)\left(8x+12+15x-6\right)\\ =\left(-7x+18\right)\left(23x+6\right)\)