Question

[kudoshinichi]

Giúp e với ah

Answer 1

[Nguyễn Hoàng Minh]

\(1,=\dfrac{x+3-x}{x\left(x+3\right)}=\dfrac{3}{x\left(x+3\right)}\\ 2,=\dfrac{20x+2x-10}{2x-5}=\dfrac{22x-10}{2x-5}\\ 3,=\dfrac{x+5+x-7+x-4}{x-2}=\dfrac{3\left(x-2\right)}{x-2}=3\\ 4,=\dfrac{x^2+3x-x^2-2x-1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\\ 5,=\dfrac{3xy+y^3-3xy-x^3}{x^2y^2}=\dfrac{y^3-x^3}{x^2y^2}\\ 6,=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)

\(7,=\dfrac{x^2+6x-2x+4}{x\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x\left(x-2\right)}\\ 8,=\dfrac{4x^2+8x+3x-6-32}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2+11x-38}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)\left(4x+19\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x+19}{x+2}\\ 9,=\dfrac{-5x-1-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{-\left(5x-1\right)^2}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{1-5x}{x\left(5x+1\right)}\\ 10,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x-4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}=\dfrac{1}{x}\)

Answer 2

[kudoshinichi]

Em camon ă