Question

[Nguyễn Thị Minh Huyền]

bài tập. Giải các pt

1, \(\frac{5}{x-2}+\frac{6}{3-4x}=0\)

2,\(\frac{x+1}{x-2}=\frac{1}{x^2-4}\)

3,\(\frac{x+2}{x}-\frac{x^2+5x+4}{x\left(x+2\right)}=\frac{x}{x+2}\)

4,\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

5,\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

6,\(\frac{x+1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)

7,\(\frac{2}{x+1}-\frac{3x+1}{\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x-2\right)}\)

8,\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)

9,\(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\)

Answer 1

[๖²⁴ʱ๖ۣۜTɦủү❄吻༉]

ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)

\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)

\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)

\(15-20x+6x-12=0\)

\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn