Question

Bài tập: Giải các phương trình sau:

1) \(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

2) \(\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}=\dfrac{5-x}{4x^2-8x}+\dfrac{7}{8x}\)

3) \(\left(x^2-4\right)-\left(x-2\right)\left(4x-5\right)=0\)

4) \(\left|-2x\right|=3x-7\)

5) \(\left|2x+1\right|-\left|5x-2\right|=3\)

6) \(\left|3x-1\right|=\left|5x+2\right|\)

Answer 1

`1)`\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

\(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x^2-25\right)}\)

\(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\)

\(ĐK:x\ne0;\pm5\)

\(\Leftrightarrow\dfrac{2\left(x+5\right)^2-\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)

\(\Leftrightarrow x^2+5x-25=0\)

\(\Delta=5^2-4.\left(-25\right)=25+100=125>0\)

`->` pt có 2 nghiệm

\(\left\{{}\begin{matrix}x=\dfrac{-5+\sqrt{125}}{2}=\dfrac{-5+5\sqrt{5}}{2}\\x=\dfrac{-5-\sqrt{125}}{2}=\dfrac{-5-5\sqrt{5}}{2}\end{matrix}\right.\) \((tm)\)

Vậy \(S=\left\{\dfrac{-5\pm5\sqrt{5}}{2}\right\}\)

`2)`\(\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}=\dfrac{5-x}{4x^2-8x}+\dfrac{7}{8x}\)

\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}=\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}\)

\(ĐK:x\ne0;2\)

\(\Leftrightarrow\dfrac{4\left(x-1\right)+x}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)+7\left(x-2\right)}{8x\left(x-2\right)}\)

\(\Leftrightarrow4x-4+x=10-2x+7x-14\)

\(\Leftrightarrow0=0\left(đúng\right)\)

Vậy pt có vô số nghiệm

Answer 2

`3)`\(\left(x^2-4\right)-\left(x-2\right)\left(4x-5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(4x-5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-4x+5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy \(S=\left\{2;\dfrac{7}{3}\right\}\)

`4)`\(\left|-2x\right|=3x-7\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=3x-7;x\le0\\2x=3x-7;x>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-5x=-7\\-x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\left(ktm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{7\right\}\)