1.

7x(2x-1)=14x²-7x

2

a. x²+2x=x(x+2)

b.x²-xy+3x-3y

=x(x-y)+3(x-y)

=(x+3)(x-y)

Câu 2:

 1. 2x/2x-5 -  5/2x-5

=2x-5/2x-5

=1

2. (6x³-7x²-x+2) : (x-1)=6x²-x-2

3x²−5x+2 dư 0

(x²– y² + 6x + 9) : (x + y + 3) = (x² + 6x+ 9) – y² : (x + y + 3)

=(x + 3)² – y² : (x + y + 3) = (x + 3 – y) (x + 3 + y) : (x + y + 3) = (x – y + 3)

còn câu c thì sao bn

Bạn hãy làm 1 điều coa thể

ok,tui nhất trí với bn nguyễn trọng bằng

\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)

\(=\left(x+\frac{1}{2}\right)\left(x-1\right)\left(x-\frac{2}{3}\right):\left(x+\frac{1}{2}\right)\)

\(=\left(x-1\right)\left(x-\frac{2}{3}\right)\)

a) Ta có: 6x^3 - 7x² - x+2 = 6x³+3x²-10x²-5x+4x+2

                                      = 3x² ( 2x+1) - 5x(2x+1) + 2(2x+1)

                                      = (2x+1)(3x²-5x+2)

Suy ra: (6x³-7x²-x+2): (2x+1)= 3x²-5x+2

b) Ta có: x² -y²+6x+9= (x²+6x+9) - y²

                               = (x+3)² - y²

                               = (x+3-y)(x+3+y)

Suy ra: (x²-y²+6x+9): (x+y+3)= x-y+3

Làm vậy nha pạn :) 

nếu tao biết tao chết liền

a.

3: \(\Leftrightarrow a-15=0\)

hay a=15

=> (x^3-2x^2+x-1)(x-5) = x^4-7x^3+11x^2-6x+5

câu 1:

x³-1+3x²-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)

Câu 2 :

a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)

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

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

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

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

b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)

\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)

Câu 3 :

Sửa đề :

\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)