giúp mình với ạ câu nào cũng được

a) \(\dfrac{x+5}{3\left(x-1\right)}+1=\dfrac{3x+7}{5\left(x-1\right)}\) ( đk: \(x\ne1\))

\(\Leftrightarrow\dfrac{5\left(x+5\right)}{15\left(x-1\right)}+\dfrac{15\left(x-1\right)}{15\left(x-1\right)}=\dfrac{3\left(3x+7\right)}{15\left(x-1\right)}\)

\(\Rightarrow5\left(x+5\right)+15\left(x-1\right)=3\left(3x+7\right)\)

\(\Leftrightarrow5x+25+15x-15=9x+21\)

\(\Leftrightarrow5x+15x-9x=21-25+15\)

\(\Leftrightarrow11x=11\Leftrightarrow x=1\) (loại)

Vậy tập nghiệm: \(S=\varnothing\)

b) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}-\dfrac{8}{x^2+2x-3}=1\)  (đk: \(x\ne1,x\ne-3\))

\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\dfrac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}-\dfrac{8}{x^2+2x-3}=\dfrac{x^2+2x-3}{x^2+2x-3}\)

\(\Rightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-8=x^2+2x-3\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5-8=x^2+2x-3\)

\(\Leftrightarrow3x=3\Leftrightarrow x=1\) (loại)

Vậy tập nghiệm: \(S=\varnothing\)

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

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

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

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

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

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)

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

Suy ra: \(2x+5-3x+6=6x-9\)

\(\Leftrightarrow-x+11-6x+9=0\)

\(\Leftrightarrow20-7x=0\)

\(\Leftrightarrow7x=20\)

hay \(x=\dfrac{20}{7}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

A 1-2x/4-1<1-6x/8

<=>2(1-2x)-8<1-6x

<=>2-4x-8<1-6x

<=>-4x+6x<1-2+8

<=>2x<7

<=>x<7/2

a)\(\dfrac{1-2x}{4}-1< \dfrac{1-6x}{8}\)

\(\Leftrightarrow\dfrac{1-2x-4}{4}< \dfrac{1-6x}{8}\)

\(\Leftrightarrow8\left(-3-2x\right)< 4\left(1-6x\right)\)

\(\Leftrightarrow-24+16x< 4-24x\)

\(\Leftrightarrow40x< 28\)

\(\Leftrightarrow x< \dfrac{7}{10}\)

b)\(\dfrac{x-1}{3}-2x>\dfrac{3x+1}{2}\)

\(\Leftrightarrow\dfrac{x-1-6x}{3}>\dfrac{3x+1}{2}\)

\(\Leftrightarrow2\left(-5x-1\right)>3\left(3x+1\right)\)

\(\Leftrightarrow-10x-2>9x+3\)

\(\Leftrightarrow-19x>5\)

\(\Leftrightarrow x< \dfrac{-5}{19}\)

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)