1/ a/ 5x² - 20

= 5.(x² - 4)

=5.(x² - 2²)

=5.(x+2).(x-2)

b/ xy² - y³ - x + y

= (xy² - x) - (y³ - y)

= x(y² - 1) - y(y² - 1)

= (y² - 1).(x-y)

= (y-1).(y+1).(x-y)

c/ x² + 3x - 10

= x² + 5x - 2x - 10

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

= (x+5).(x-2)

d/ x² - y² + 12y - 36

= x² - (y² - 2.y.6 + 6²)

= x² - (y-6)²

= (x+y-6).(x-y+6).

2/ a/ 4x² - 9 - x(2x-3) = 0

(2x)² - 3² - x(2x-3) = 0

(2x+3).(2x-3)-x(2x-3) = 0

(2x-3).(2x+3-x) = 0

(2x-3).(x+3) = 0

=> 2x - 3 = 0 hoặc x + 3 = 0

hay x = 3/2 hoặc x = -3

b/ x³ -25x = 0

x(x² - 25) = 0

x(x+5)(x-5) = 0

=> x = 0 hoặc x+5=0 hoặc x-5 = 0

hay x = 0; x = -5; x = 5

c/ 2(x+5) - x² - 5x = 0

2(x+5) - x(x+5) = 0

(x+5).(2-x) = 0

=> x + 5 = 0 hoặc 2 - x = 0

hay x = -5 hoặc x = 2

d/ 2x² + 5x - 3 = 0

2x² - x + 6x - 3 = 0

x(2x-1) + 3(2x-1) = 0

(2x-1).(x+3) = 0

=> 2x-1=0 hoặc x+3=0

hay x = 1/2 hoặc x = -3

Ai giúp mk với mk đang cần gấp

Mk làm được hết

mà vẫn cứ sai hoài à

tìm mãi ko thấy lỗi sai

Bài 3  Tính nhanh

A, 892^2+892.216+108^2                 B, 36^2+26^2-52.36

  =892^2+2.892.108+108^2                 =36^2-52.62+26^2

  =(892+108)^2           

  =1000^2

  =1000000

Bài 4 Phân tích đa thức sau thành nhân tử   

X^3-2x^2+x

5(x-y)-y(x-y)

36-12x+x^2

4x^2+12x-9

Bài 3:

\(892^2+892.216+108^2=892^2+2.892.108+108^2=\left(892+108\right)^2=1000000\)

\(36^2+26^2-52.36=36^2-2.26.36+26^2=\left(36-26\right)^2=100\)

Bài 4: 

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

\(5.\left(x-y\right)-y.\left(x-y\right)=\left(5-y\right)\left(x-y\right)\)

\(36-12x+x^2=x^2-12x+36=x^2-2x.6+6^2=\left(x-6\right)^2\)

\(4x^2+12x-9=\left(2x\right)^2+2.2x.3+3^2=\left(2x+3\right)^2\)

?1 . Có . Mẫu thức chung : 12x²y³z đơn giản hơn

?2 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)

\(\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)

?3 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)

\(\dfrac{-5}{10-2x}=\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)

a: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(2-x\right)\left(2+x\right)}\)

\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=\dfrac{8x+16}{x\left(2-x\right)\left(2+x\right)}\)

b: \(x^2+1=\dfrac{x^4-1}{x^2-1}\)

\(\dfrac{x^4}{x^2-1}=\dfrac{x^4}{x^2-1}\)

c: \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3y}{y\cdot\left(x-y\right)^3}\)

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

Bài giải

a) \(\dfrac{1}{x+2}=\dfrac{x.\left(x-2\right)}{\left(x+2\right)\left(x-2\right).x}=\dfrac{x^2-2x}{x\left(x+2\right)\left(x-2\right)}\)

\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)

b) \(x^2+1=\dfrac{x^2+1}{1}=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x^2-1}=\dfrac{x^4-1}{x^2-1}\)

\(\dfrac{x^4}{x^2-1}\) giữ nguyên.

c) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)

\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)

\(\left(x+5\right)\left(x^2-5x+25\right)\)

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

\(=x^3+5^3\)

\(=x^3+125\)

3) \(27-y^3\)

\(=3^3-y^3\)

\(=\left(3-y\right)\left(9-3y+y^2\right)\)