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

b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)

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

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

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

=1/3

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

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

c: \(=x^2+6xy+9y^2\)

e: \(=x^4-4y^2\)

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

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

\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)

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

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)

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

\(\dfrac{xy\left(x^2+y^2\right)}{xy\left(x^3\right)}.\dfrac{1}{x^2+y^2}=\dfrac{1}{x^3}\)

Bài 3:

a: \(\frac{x}{x-3}+\frac{9-6x}{x^2-3x}\)

\(=\frac{x}{x-3}+\frac{-6x+9}{x\left(x-3\right)}\)

\(=\frac{x^2-6x+9}{x\left(x-3\right)}=\frac{\left(x-3\right)^2}{x\left(x-3\right)}=\frac{x-3}{x}\)

b: \(\frac{6x-3}{x}:\frac{4x^2-1}{3x^2}\)

\(=\frac{3\left(2x-1\right)}{x}\cdot\frac{3x^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{3\cdot3x}{2x+1}=\frac{9x}{2x+1}\)

Bài 2:

a: \(\frac{x^3-x}{3x+3}\)

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

b: \(\frac{x^2+3xy}{x^2-9y^2}=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\frac{x}{x-3y}\)

Bài 1:

a: \(\frac{x^2-9}{2x+6}:\frac{3-x}{2}\)

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

b: \(\frac{2x}{x-y}-\frac{2y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)

c: \(\frac{x+15}{x^2-9}+\frac{2}{x+3}\)

\(=\frac{x+15}{\left(x-3\right)\left(x+3\right)}+\frac{2}{x+3}\)

\(=\frac{x+15+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x+15+2x-6}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}\)

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

d: \(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{y^2+x^2}{y^2-x^2}\)

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

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

\(=\frac{2x^2+4xy+2y^2}{2\left(x-y\right)\left(x+y\right)}=\frac{2\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\frac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\frac{x+y}{x-y}\)

Tham khảo

a)

-7x²(3x - 4y)

= -7x².3x + 7x^2ư.4y

= -21x² + 28x²y

b)

(x - 3)(5x - 4)

= x.5x - x.4 - 3.5x + 3.4

= 5x² - 4x - 15x + 12

= 5x² - 19x + 12

c)

(2x - 1)² = 4x² - 4x + 1

d)

(x + 3)(x - 3) = x² - 3² = x² - 9

a, = -21x³ +28x²y

b, = 5x² -4x -15x +12

= 5x² -19x +12

c, =x² -9

\(a,=-21x^3+28x^2y\\ b,=5x^2-4x-15x+12=5x^2-19x+12\\ c,=4x^2-4x+1\\ d,=49-x^2\)

\(a,=2x^2-7xy-30y^2+30x^2+2xy=32x^2-5xy-30y^2\\ b,=x^2-10x+25+2x^2-8=3x^2-10x+17\\ c,=x^3+8-x^3+5=13\\ d,=x^3-x^2+x-x^2+x-1+x^2-1=x^3-x^2+2x-2\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4: