Question

[Garuda]

Bài1: giải các phương trình sau: A) \(\frac{1}{x+1}+\frac{2}{x-1}=\frac{1+xmũ2}{xmũ2-1}\) B) \(\frac{X-2}{x+2}-\frac{X}{x-2}=\frac{8}{xmũ2-4}\) C) \(\frac{1}{x}+\frac{2}{x-3}=\frac{1-5x}{xmũ2-3x}\) Bài2: giải các pt sau: A)\(\frac{1}{x+2}=\frac{5}{2-x}+\frac{12+x}{xmũ2-4}\) B) \(\frac{1}{x+4}=\frac{5}{4-x}-\frac{3+x}{Xmũ2-16}\)

Answer 1

[Trương Huy Hoàng]

Bài 1:

a, \(\frac{1}{x+1}+\frac{2}{x-1}=\frac{1+x^2}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)

\(\Leftrightarrow\) \(\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{1+x^2}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow\) x - 1 + 2(x + 1) = 1 + x²

\(\Leftrightarrow\) x - 1 + 2x + 2 - 1 - x² = 0

\(\Leftrightarrow\) -x² + 3x = 0

\(\Leftrightarrow\) x(3 - x) = 0

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

Vậy S = {0; 3}

b, \(\frac{x-2}{x+2}-\frac{x}{x-2}=\frac{8}{x^2-4}\) (ĐKXĐ: x \(\ne\) \(\pm\) 2)

\(\Leftrightarrow\) \(\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow\) (x - 2)² - x(x + 2) = 8

\(\Leftrightarrow\) (x - 2)² - x(x + 2) - 8 = 0

\(\Leftrightarrow\) x² - 4x + 4 - x² - 2x - 8 = 0

\(\Leftrightarrow\) -6x - 4 = 0

\(\Leftrightarrow\) x = \(\frac{-2}{3}\) (TMĐKXĐ)

Vậy S = {\(\frac{-2}{3}\)}

c, \(\frac{1}{x}\) + \(\frac{2}{x-3}\) = \(\frac{1-5x}{x^2-3x}\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) 3)

\(\Leftrightarrow\) \(\frac{x-3}{x\left(x-3\right)}+\frac{2x}{x\left(x-3\right)}=\frac{1-5x}{x\left(x-3\right)}\)

\(\Rightarrow\) x - 3 + 2x = 1 - 5x

\(\Leftrightarrow\) 3x - 3 = 1 - 5x

\(\Leftrightarrow\) 3x + 5x = 1 + 3

\(\Leftrightarrow\) 8x = 4

\(\Leftrightarrow\) x = \(\frac{1}{2}\) (TMĐKXĐ)

Vậy S = {\(\frac{1}{2}\)}

Bài 2:

a, \(\frac{1}{x+2}=\frac{5}{2-x}+\frac{12+x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{1}{x+2}=\frac{-5}{x-2}+\frac{12+x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\) \(\frac{x-2}{\left(x+2\right)\left(x-2\right)}=\frac{-5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12+x}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow\) x - 2 = -5(x + 2) + 12 + x

\(\Leftrightarrow\) x - 2 = -5x - 10 + 12 + x

\(\Leftrightarrow\) x - 2 = -4x + 2

\(\Leftrightarrow\) x + 4x = 2 + 2

\(\Leftrightarrow\) 5x = 4

\(\Leftrightarrow\) x = \(\frac{4}{5}\)

Vậy S = {\(\frac{4}{5}\)}

Chúc bn học tốt!! (Phần b hình như không có gì thì phải)