Bài1: giải các phương trình sau: 1)\(\frac{2x-5}{x+5}=3\) 2)\(\frac{4}{x+1}=\frac{3}{x-2}\) 3) \(\frac{5}{2x-3}=\frac{1}{x-4}\) Bài2: giải các phương trình sau: 1)\(\frac{1}{x-1}+\frac{2}{x+1}=\frac{5x-3}{xmũ2-1}\) 2) \(\frac{x+2}{x-2}-\frac{1}{X}=\frac{2}{xmũ2-2x}\) 3) \(\frac{5}{x-3}-\frac{3}{x+3}=\frac{3x}{xmũ2-9}\)
Bài 1:
1, \(\frac{2x-5}{x+5}=3\) (ĐKXĐ: x \(\ne\) -5)
\(\Leftrightarrow\) \(\frac{2x-5}{x+5}=\frac{3\left(x+5\right)}{x+5}\)
\(\Rightarrow\) 2x - 5 = 3(x + 5)
\(\Leftrightarrow\) 2x - 5 = 3x + 15
\(\Leftrightarrow\) 2x - 3x = 15 + 5
\(\Leftrightarrow\) -x = 20
\(\Leftrightarrow\) x = -20 (TMĐKXĐ)
Vậy S = {-20}
2, \(\frac{4}{x+1}=\frac{3}{x-2}\) (ĐKXĐ: x \(\ne\) -1; x \(\ne\) 2)
\(\Leftrightarrow\) \(\frac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\) 4(x - 2) = 3(x + 1)
\(\Leftrightarrow\) 4x - 8 = 3x + 3
\(\Leftrightarrow\) 4x - 3x = 3 + 8
\(\Leftrightarrow\) x = 11 (TMĐKXĐ)
Vậy S = {11}
3, \(\frac{5}{2x-3}=\frac{1}{x-4}\) (ĐKXĐ: x \(\ne\) \(\frac{3}{2}\); x \(\ne\) 4)
\(\Leftrightarrow\) \(\frac{5\left(x-4\right)}{\left(2x-3\right)\left(x-4\right)}=\frac{2x-3}{\left(2x-3\right)\left(x-4\right)}\)
\(\Rightarrow\) 5(x - 4) = 2x - 3
\(\Leftrightarrow\) 5x - 20 = 2x - 3
\(\Leftrightarrow\) 5x - 2x = -3 + 20
\(\Leftrightarrow\) 3x = 17
\(\Leftrightarrow\) x = \(\frac{17}{3}\) (TMĐKXĐ)
Vậy S = {\(\frac{17}{3}\)}
Bài 2:
1, \(\frac{1}{x-1}+\frac{2}{x+1}=\frac{5x-3}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)
\(\Leftrightarrow\) \(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{5x-3}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow\) x + 1 + 2(x - 1) = 5x - 3
\(\Leftrightarrow\) x + 1 + 2x - 2 = 5x - 3
\(\Leftrightarrow\) 3x - 1 = 5x - 3
\(\Leftrightarrow\) 3x - 5x = -3 + 1
\(\Leftrightarrow\) -2x = -2
\(\Leftrightarrow\) x = 1 (KTM)
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
2, \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\) (ĐKXĐ: x \(\ne\) 2; x \(\ne\) 0)
\(\Leftrightarrow\) \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Rightarrow\) x(x + 2) - x + 2 = 2
\(\Leftrightarrow\) x2 + 2x - x + 2 = 2
\(\Leftrightarrow\) x2 + x = 2 - 2
\(\Leftrightarrow\) x2 + x = 0
\(\Leftrightarrow\) x(x + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 0 và x = -1
Ta có: x = 0 KTM đkxđ
\(\Rightarrow\) x = -1
Vậy S = {-1}
3, \(\frac{5}{x-3}-\frac{3}{x+3}=\frac{3x}{x^2-9}\) (ĐKXĐ: x \(\ne\) \(\pm\) 3)
\(\Leftrightarrow\) \(\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\) 5(x + 3) - 3(x - 3) = 3x
\(\Leftrightarrow\) 5x + 15 - 3x + 9 = 3x
\(\Leftrightarrow\) 2x + 24 = 3x
\(\Leftrightarrow\) 2x - 3x = 24
\(\Leftrightarrow\) -x = 24
\(\Leftrightarrow\) x = -24 (TMĐKXĐ)
Vậy S = {-24}
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