Question

[Bach Tang Oni]

Bài 1:

i)\(\dfrac{x+1}{x-5}\)+\(\dfrac{x-18}{x-5}\)-\(\dfrac{x+2}{5-x}\)

j)\(\dfrac{3x\left(x-2\right)}{3x-2}\)+\(\dfrac{6x^2}{3x-2}\)-\(\dfrac{2\left(2-3x\right)}{2-3x}\)

n)\(\dfrac{2}{x}\)+\(\dfrac{3}{x-1}\)+\(\dfrac{1-4x}{x^2-x}\)

Bài 2:

j)\(\dfrac{2}{3x}\)-\(\dfrac{1}{2x-2}\)-\(\dfrac{x-4}{6x-6x^2}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

i: \(=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)

Answer 2

[ILoveMath]

Bài 1:

\(i,\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}-\dfrac{x+2}{5-x}=\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)

\(j,\dfrac{3x\left(x-2\right)}{3x-2}+\dfrac{6x^2}{3x-2}-\dfrac{2\left(2-3x\right)}{2-3x}=\dfrac{3x^2-6x}{3x-2}+\dfrac{6x^2}{3x-2}+\dfrac{4-6x}{3x-2}=\dfrac{3x^2-6x+6x^2+4-6x}{3x-2}=\dfrac{9x^2-12x+4}{3x-2}=\dfrac{\left(3x-2\right)^2}{3x-2}=3x-2\)

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

Bài 2:

\(j,\dfrac{2}{3x}-\dfrac{1}{2x-2}-\dfrac{x-4}{6x-6x^2}=\dfrac{4\left(x-1\right)}{6x\left(x-1\right)}-\dfrac{3x}{6x\left(x-1\right)}-\dfrac{x-4}{6x\left(1-x\right)}=\dfrac{4x-4-3x+x-4}{6x\left(x-1\right)}=\dfrac{2x-8}{6x\left(x-1\right)}=\dfrac{2\left(x-4\right)}{6x\left(x-1\right)}=\dfrac{x-4}{3x\left(x-1\right)}\)