K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

17 tháng 7 2018

d) mk chỉnh lại đề

  \(8xy^2-5xyz-24y+15z\)

\(=xy\left(8y-5z\right)-3\left(8y-5z\right)\)

\(=\left(8y-5z\right)\left(xy-3\right)\)

e)   \(x^4-x^3-x+1=\left(x-1\right)^2\left(x^2+x+1\right)\)

f)  \(x^4+x^2y^2+y^4=\left(x^2-xy+y^2\right)\left(x^2+xy-y^2\right)\)

g)  \(x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)\)

h)   \(x^3-3x^2+2=\left(x-1\right)\left(x^2-2x-2\right)\)

i)  \(2x^3+x^2-4x-12=\left(x-2\right)\left(2x^2+5x+6\right)\)

k)  \(25x^2\left(x-5\right)-x+y=\left(1-5x\right)\left(1+5x\right)\left(y-x\right)\)

18 tháng 6 2016

a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)

b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)

c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)

d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2

= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)

e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)

f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)

g) chắc là 3xyz 

= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)

h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)

i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy

k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).

28 tháng 7 2017

a, \(x^3-x^2y-xy^2+y^3\)

\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)

\(=\left(x^2-y^2\right)\left(x-y\right)\)

\(=\left(x-y\right)^2\left(x+y\right)\)

b, \(x^3+x^2-4x-4\)

\(=x^2\left(x+1\right)-4\left(x+1\right)\)

\(=\left(x^2-4\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)

c, \(x^3-x^2-x+1\)

\(=x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)\)

\(=\left(x-1\right)^2\left(x+1\right)\)

d, \(\left(7x+3\right)^2-\left(2x-1\right)^2\)

\(=\left(7x+3-2x+1\right)\left(7x+3+2x-1\right)\)

\(=\left(5x+4\right)\left(9x+2\right)\)

e, \(x^3-3x^2-3x+1\) sai đề

f, \(x^2-2x-3\)

\(=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)\)

\(=\left(x+1\right)\left(x-3\right)\)

g, \(x^2-2x-8\)

\(=x^2-4x+2x-8=x\left(x-4\right)+2\left(x-8\right)\)

\(=\left(x+2\right)\left(x-8\right)\)

h, \(x^2-10x+21\)

\(=x^2-7x-3x+21\)

\(=x\left(x-7\right)-3\left(x-7\right)=\left(x-3\right)\left(x-7\right)\)

i, \(x^2-4xy+3y^2\)

\(=x^2-4xy+4y^2-y^2\)

\(=\left(x-2y\right)^2-y^2\)

\(=\left(x-2y-y\right)\left(x-2y+y\right)\)

\(=\left(x-3y\right)\left(x-y\right)\)

28 tháng 7 2017

a) \(x^3 - x^2y - xy^2 + y^3\)

\(=\left(x^3-x^2y\right)-\left(xy^2-y^3\right)\)

\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-y^2\right)\)

\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)^2\left(x+y\right)\)

b) \(x^3 + x^2 - 4x - 4\)

\(=\left(x^3+x^2\right)-\left(4x+4\right)\)

\(=x^2\left(x+1\right)-4\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-4\right)\)

\(=\left(x+1\right)\left(x^2-2^2\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x-2\right)\)

4 tháng 8 2017

Mình sửa: Bài 1
2)x2+3x-15

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)

a) Ta có: \(x^2+4x+3\)

\(=x^2+x+3x+3\)

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

b) Ta có: \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15x+x-3\)

\(=-5x\left(x-3\right)+\left(x-3\right)\)

\(=\left(x-3\right)\left(-5x+1\right)\)

c) Ta có: \(2x^2+7x+5\)

\(=2x^2+2x+5x+5\)

\(=2x\left(x+1\right)+5\left(x+1\right)\)

\(=\left(x+1\right)\left(2x+5\right)\)

d) Ta có: \(2x^2+3x-5\)

\(=2x^2+5x-2x-5\)

\(=x\left(2x+5\right)-\left(2x+5\right)\)

\(=\left(2x+5\right)\left(x-1\right)\)

e) Ta có: \(x^3-3x^2+1-3x\)

\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)

\(=\left(x+1\right)\left(x^2-4x+1\right)\)

f) Ta có: \(x^2-4x-5\)

\(=x^2-4x+4-9\)

\(=\left(x-2\right)^2-3^2\)

\(=\left(x-2-3\right)\left(x-2+3\right)\)

\(=\left(x-5\right)\left(x+1\right)\)

g) Ta có: \(\left(a^2+1\right)^2-4a^2\)

\(=\left(a^2+1\right)^2-\left(2a\right)^2\)

\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)

\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)

h) Ta có: \(x^3-3x^2-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-4\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

i) Ta có: \(x^4+x^3+x+1\)

\(=x^3\left(x+1\right)+\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+1\right)\)

\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)

k) Ta có: \(x^4-x^3-x^2+1\)

\(=x^3\left(x-1\right)-\left(x^2-1\right)\)

\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^3-x-1\right)\)

l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)

\(=3x\left(x+2\right)\)

m) Ta có: \(x^4+4x^2-5\)

\(=x^4-x^2+5x^2-5\)

\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2+5\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

1 tháng 9 2018

Cái này chưa học bt làm mấy câu

b. x^2 + 2x - 3

= x^2 + 3x - x - 3

= x ( x - 1 ) + 3 ( x - 1 )

= ( x + 3 ) ( x - 1 )

1 tháng 9 2018

\(4x^2-3x-4\)

\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)

\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)

\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)

\(x^2+2x-3\)

\(=x^2-x+3x-3\)

\(=x\left(x-1\right)+3\left(x-1\right)\)

\(=\)\(\left(x+3\right)\left(x-1\right)\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)

đặt \(x^2+5x+5=t\)

\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)

            \(=t^2-1-24\)

            \(=t^2-25\)

            \(=\left(t-5\right)\left(t+5\right)\)

hay \(\left(1\right)=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)

               \(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

                \(=x\left(x+5\right)\left(x^2+5x+10\right)\)

học tốt

17 tháng 6 2015

c, =(5x)^3 + (y^2)^ 3 = (5x+y^2)(25x^2 - 5xy^2 + y^4)

d, = (0,5.(a+1))^3-1^3 = ( 0,5(a+1) - 1 ) ( 0,25(a+1) ^2 +a,5(a+1) + 1)

e,2x( x+ 1 ) + 2(x+ 1 ) = 2(x+1)(x+1) = 2(x+1)^2

g, y^2 (x^2 + y) - zx^2 - zy = x^2.y^2 - z.x^2 + y^3 - zy = x^2 (y^2 - z) + y (y^2 -z) = (x^2 +y) (y^2 -z)

h,4.x(x-2y) + 8.y(2y -x) = 4x( x- 2 y ) -8 (x - 2y) = (4x - 8) (x-2y)=4(x-2)(x-2y)

k,=(x+1)(3x(x+1)-5x+7) =(x+1) (3x^2 +3x - 5x + 7)