b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

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

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

help me pls!!!

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

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

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

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}

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

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

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

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

\(\Leftrightarrow7+4x=15\)

\(\Leftrightarrow x=2\)

Vậy x = 2

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

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

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

\(\Leftrightarrow\dfrac{3x+6-6+2x}{\left(x+2\right)\left(3-x\right)}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}\)

\(\Leftrightarrow5x=5x\)

\(\Leftrightarrow x\in R\)

Vậy...

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

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

\(\Leftrightarrow\dfrac{18-6x+4x-4}{\left(x-1\right)\left(3-x\right)}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)

\(\Leftrightarrow14-2x=8\)

\(\Leftrightarrow x=3\)

Vậy x = 3

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

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

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

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

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

Vậy...

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)