\(\dfrac{5x+5y}{3x-3y}:\dfrac{5x}{x^2-y^2}.\)

\(=\dfrac{5\left(x+y\right)}{3\left(x-y\right)}.\dfrac{\left(x-y\right)\left(x+y\right)}{5x}.\)

\(=\dfrac{x+y}{3}.\dfrac{x+y}{x}.\)

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

Cảm ơn nha

b: \(=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

\(a,=-3x^3+x^2+9x^2-3x-12x+4=-3x^3+10x^2-15x+4\\ b,=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

\(\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}.\left(x\ne1\right).\)

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

a) 4x²y³.\(\dfrac{2}{4}\)x³y

= (4.\(\dfrac{2}{4}\))(x².x³)(y³.y)

=\(\dfrac{1}{2}\)x⁵y⁴

b)(5x-2)(25x²+10x+4)

=(5x-2)(5x+2)²

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

=(5x²-2²)(5x+2)

a) 4x²y³ .2/4x³y

= 2x⁵y⁴

b) (5x-2)(25x²+10x+4)

= 125x³+50x²+20x-50x²-20x-8.

= 125x³-8

1: ĐKXĐ:  \(x\notin\left\{0;3\right\}\)

\(\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\)

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

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

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

=0

2: ĐKXĐ: \(x\notin\left\{0;1\right\}\)

\(\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x^2-x}\)

\(=\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x\left(x-1\right)}\)

\(=\dfrac{3x-3-5x+3x+2}{x\left(x-1\right)}\)

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

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

thanks

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(3x+9+4x-12=3x-7\)

\(\Leftrightarrow4x=-7+12-9=-4\)

hay \(x=-1\left(nhận\right)\)

2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)

\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(3x+12-4x+16=3x-4\)

\(\Leftrightarrow28-4x=-4\)

\(\Leftrightarrow4x=32\)

hay \(x=8\left(tm\right)\)

3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

Suy ra: \(5x^2-12+3x+3=5x^2-5x\)

\(\Leftrightarrow3x-9+5x=0\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(nhận\right)\)