Question

a) x +\(\frac{5}{x}\)> 0

b) \(\frac{6}{x^2-1}\)+ 5 = \(\frac{8x-1}{4x+4}-\frac{12-1}{4-4x}\)

c) (x²  - 2x) (x³ -3x²-18x) =0

d) \(\frac{x+1}{12}-\frac{x-1}{6}>\frac{x-2}{8}-\frac{x+3}{8}\)

e) / 2x-3/ = x-1

f) /x-5/-5=7

Answer 1

\(e)\) \(\left|2x-3\right|=x-1\)

Ta có : 

\(\left|2x-3\right|\ge0\)\(\left(\forall x\inℚ\right)\)

Mà \(\left|2x-3\right|=x-1\)

\(\Rightarrow\)\(x-1\ge0\)

\(\Rightarrow\)\(x\ge1\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-3=x-1\\2x-3=1-x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-x=-1+3\\2x+x=1+3\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\3x=4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=\frac{4}{3}\left(tm\right)\end{cases}}}\)

Vậy \(x=2\) hoặc \(x=\frac{4}{3}\)

Chúc bạn học tốt ~ 

Answer 2

\(f)\) \(\left|x-5\right|-5=7\)

\(\Leftrightarrow\)\(\left|x-5\right|=12\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=12\\x-5=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=17\\x=-7\end{cases}}}\)

Vậy \(x=17\) hoặc \(x=-7\)

Chúc bạn học tốt ~