Question

[Đỗ Linh Chi]

Giải phương trình:

a)\(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)

b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\)

c)\(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)

d) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

Answer 1

[Võ Đông Anh Tuấn]

a ) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)(1)

ĐKXĐ : \(x\ne1;x\ne2\)

(1)\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Leftrightarrow2-x+5x+5=15\)

\(\Leftrightarrow4x+7=15\\\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\left(KTMĐKXĐ\right)\)

Vậy pt vô nghiệm .

b ) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( 2 )

ĐKXĐ : \(x\ne3;x\ne-2\)

(2) \(\Leftrightarrow3x-x^2+6-2x+x^2+2x=3x+6-x^2-2x\)

\(\Leftrightarrow x^2+2x=0\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-2\left(KTMĐKXĐ\right)\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là S={0}.

c ) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) (3)

ĐKXĐ : \(x\ne1;x\ne3\)

\(\left(3\right)\Leftrightarrow\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)

\(\Leftrightarrow6\left(3-x\right)+4\left(x-1\right)=8\)

\(\Leftrightarrow18-6x+4x-4=8\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow x=-3\)

Vậy tập nghiệm của phương trình là S={-3}

d ) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (4)

ĐKXĐ : \(x\ne0;x\ne2\)

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

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là S={-1}

Answer 2

[Rain Tờ Rym Te]

a) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) ( đk: x ≠ -1; x ≠ 2 )

\(\Leftrightarrow\) \(\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Leftrightarrow\) \(2-x+5\left(x+1\right)=15\)

\(\Leftrightarrow\) \(2-x+5x+5=15\)

\(\Leftrightarrow\)\(4x=8\)

\(\Rightarrow\) \(x=2\) ( KTM )

S = ∅

b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( đk: x ≠ - 2 ; x ≠ 3 )

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

\(\Leftrightarrow\) \(3x-x^2+6-2x+x^2+2x=5x+6-2x\)

\(\Leftrightarrow\) \(3x+6=3x+6\)

\(\Rightarrow\)\(0x=0\) ( TM )

\(\Rightarrow\) Phương trình vô số nghiệm

S = R

c) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) ( đk: x ≠ 1 ; x ≠ 3 )

\(\Leftrightarrow\) \(\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)

\(\Leftrightarrow\)\(6\left(3-x\right)+4\left(x-1\right)=8\)

\(\Leftrightarrow\) \(18-6x+4x-4=8\)

\(\Leftrightarrow\) \(-2x=-6\)

\(\Rightarrow x=3\) ( KTM )

S = ∅

d) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (đk: x ≠ 2; x ≠ 0 )

\(\Leftrightarrow\) \(x\left(x+2\right)-x+2=2\)

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

\(\Leftrightarrow\) \(x^2+x=0\)

\(\Leftrightarrow\) \(x\left(x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)

S = \(\left\{2\right\}\)