Question

Giải các phương trình và bất phương trình sau

a)\(\left|x-9\right|\) \(=2x+5\)

b) \(\dfrac{1-2x}{4}\) \(-2\) ≤ \(\dfrac{1-5x}{8}\) + x

c)\(\dfrac{2}{x-3}\)\(+\dfrac{3}{x+3}\)\(=\dfrac{3x+5}{x^2-9}\)

Answer 1

|x-9|=2x+5

Xét 3 TH

TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)

TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)

TH3: x=9 =>0=23(L)

Vậy  x= 4/3

Answer 2

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

\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)

\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)

Answer 3

Ta có:

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)

\(\dfrac{2\left(x+3\right)+3\left(x-3\right)}{x^2-9}=\dfrac{3x+5}{x^2-9}\)

\(5x-4=3x+5\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\)

Answer 4

a) Ta có: |x-9|=2x+5

\(\Leftrightarrow\left[{}\begin{matrix}x-9=2x+5\left(x\ge9\right)\\9-x=2x+5\left(x< 9\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-9-2x-5=0\\9-x-2x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x-14=0\\-3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=14\\-3x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-14\left(loại\right)\\x=\dfrac{4}{3}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{4}{3}\right\}\)