Question

[TRịnh Kim Ngân]

 

Answer 1

[HT.Phong (9A5)]

Bài 2:

\(a.\dfrac{5x-3}{x+2}-4=\dfrac{6}{x+2}\left(x\ne-2\right)\\ \Leftrightarrow\dfrac{5x-3}{x+2}-\dfrac{6}{x+2}=4\\ \Leftrightarrow\dfrac{5x-3-6}{x+2}=4\\ \Leftrightarrow\dfrac{5x-9}{x+2}=4\\ \Leftrightarrow5x-9=4\left(x+2\right)\\ \Leftrightarrow5x-9=4x+8\\ \Leftrightarrow5x-4x=8+9\\ \Leftrightarrow x=17\left(tm\right)\)

\(b.\dfrac{1}{x}-\dfrac{x+2}{x-2}=\dfrac{-2}{x\left(x-2\right)}\left(x\notin\left\{0;2\right\}\right)\\ \Leftrightarrow\dfrac{x-2}{x\left(x-2\right)}-\dfrac{x\left(x+2\right)}{x\left(x-2\right)}=\dfrac{-2}{x\left(x-2\right)}\\ \Leftrightarrow x-2-x\left(x+2\right)=-2\\ \Leftrightarrow x-2-x^2-2x=-2\\ \Leftrightarrow-x^2-x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

\(c.\dfrac{7}{x-5}-2=\dfrac{3}{5-x}\left(x\ne5\right)\\ \Leftrightarrow\dfrac{7}{x-5}-2-\dfrac{3}{x-5}=0\\ \Leftrightarrow\dfrac{7}{x-5}+\dfrac{3}{x-5}=2\\ \Leftrightarrow\dfrac{10}{x-5}=2\\ \Leftrightarrow x-5=\dfrac{10}{2}=5\\\Leftrightarrow x=5+5\\ \Leftrightarrow x=10\left(tm\right)\)

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 3:

a: ĐKXĐ: \(x\notin\left\{-2;2\right\}\)

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

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

=>\(x^2-2x-\left(x^2+x-2\right)=-3x+2\)

=>-3x+2=-3x+2

=>0x=0(luôn đúng)

Vậy: \(x\in R\backslash\left\{-2;2\right\}\)

b: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

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

=>\(x^2-3x+2+x^2+3x+2-2x^2-4=0\)

=>0x=0(luôn đúng)

vậy: \(x\in R\backslash\left\{2;-2\right\}\)