\(a,2x^2+6x=2x\left(x+3\right)\\ b,x^2+2xy+y^2-9z^2\\ =\left(x^2+2xy+y^2\right)-\left(3z\right)^2\\ =\left(x+y\right)^2-\left(3z\right)^2\\ =\left(x+y-3z\right)\left(x+y+3z\right)\\ b,x^3-2x^2+x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ d,x^2-2x-15=x^2-5x+3x-15\\ =x\left(x-5\right)+3\left(x-5\right)\\ =\left(x+3\right)\left(x-5\right)\)

a) 2x² + 6x

=2x (x+3)

b) x² + 2xy +y²- 9z²

(x² + 2xy +y2)- 9z²

(x + y)² - (3z)²

(x+y-3z) (x+y+3z) 

\(=2x^2-6x+x^2-1=x^2-6x-1\)

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

\(=2x^2-6x+x^2-1\)

\(=3x^2-6x+1\)

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

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

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

b) x^2+xy/x^2-y^2= x(x+y) /(x+y) (x-y) =x/x-y

c) =2/x-1

\(C=\sqrt{3}-\sqrt{2}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{2}\\ C=\sqrt{3}+2-\sqrt{3}=2\)

5) Ta có: \(\dfrac{x^3-x^2-2x-20}{x^2-4}-\dfrac{5}{x+2}+\dfrac{3}{x-2}\)

\(=\dfrac{x^3-x^2-2x-20}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^3-x^2-2x-20-5x+10+3x+6}{\left(x-2\right)\left(x+2\right)}\)

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

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

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

\(=x-1\)

6) Ta có: \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)

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

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

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

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

\(=\dfrac{4x-1}{x^3\cdot\left(x-1\right)^2}\)

(6x+1)^2 +(6x-1)^2-2(1+6x)(6x-1)

Bài đó làm thế nào?

3/15 chia cả mẫu cả tử cho 3 được 1/5+2/5=3/5 nhé

x(x – y) + y(x + y)

= x.x – x.y + y.x + y.y

= x2 – xy + xy + y2

= x2 + y2.

Tại x = –6 ; y = 8, giá trị biểu thức bằng : (–6)2 + 82 = 36 + 64 = 100.

sai đề

x.(x2 – y) – x2.(x + y) + y.(x2 – x)

= x.x2 – x.y – (x2.x + x2.y) + y.x2 – y.x

= x3 – xy – x3 – x2y + x2y – xy

= (x3 – x3) + (x2y – x2y) – xy – xy

= –2xy

Tại  và y = –100, giá trị biểu thức bằng: