\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ ĐKXĐ:x\ne-1;x\ne2\\ \dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{1\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Rightarrow2x-4-x-1=3x-11\\ \Leftrightarrow2x-x-3x=-11+4+1\\ \Leftrightarrow-2x=-6\\ \Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

=>2x-4-x-1=3x-11

=>3x-11=x-5

=>2x=6

=>x=3

\(x\ne-1,x\ne2\\ \Leftrightarrow2x-4-x-1=3x-11\\ 6=2x\\ x=6:2=3_{\left(tmđk\right)}\)

x≠−1,x≠2⇔2x−4−x−1=3x−116=2xx=6:2=3(tmđk)

a: =>3x-9+5+10x=90

=>13x-4=90

=>13x=94

hay x=94/13

b: \(\Leftrightarrow2x-4-x-1=3x-11\)

=>3x-11=x-5

=>2x=6

hay x=3(nhận)

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.

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

\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(\Leftrightarrow x=-\dfrac{1}{12}\)

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

\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{10x+x-1}{15}=\dfrac{15}{15}-\dfrac{9x-1+2x}{15}\)

\(\Leftrightarrow15x+9x-1=14-7x\)

\(\Leftrightarrow31x=15\)

\(\Leftrightarrow x=\dfrac{15}{31}\)

câu h tương tự bài trước câu g mik làm nhé

\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)

\(ĐK:x\ne0;-10\)

\(\Leftrightarrow\dfrac{12\left(x+10\right)+12x}{12x\left(x+10\right)}=\dfrac{x\left(x+10\right)}{12x\left(x+10\right)}\)

\(\Leftrightarrow12\left(x+10\right)+12x-x\left(x+10\right)=0\)

\(\Leftrightarrow12x+120+12x-x^2-10x=0\)

\(\Leftrightarrow-x^2+14x+120=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)

\(o,\dfrac{x}{2x+6}-\dfrac{x}{2x-2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x+3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x+3\right)-2\left(3x+2\right)}{2\left(x+1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow x^2+x-x^2-3x-6x-4=0\)

\(\Leftrightarrow-8x-4=0\)

\(\Leftrightarrow-4\left(2x+1\right)=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

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

Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)

a: =>-3x=-12

=>x=4

b: =>3(3x+2)-3x-1=12x+10

=>9x+6-3x-1=12x+10

=>12x+10=6x+5

=>6x=-5

=>x=-5/6

c: =>x(x+1)+x(x-3)=4x

=>x^2+x+x^2-3x-4x=0

=>2x^2-6x=0

=>2x(x-3)=0

=>x=3(loại) hoặc x=0(nhận)

a: Ta có: \(3x-\left(3x+2\right)=x+3\)

\(\Leftrightarrow x+3=-2\)

hay x=-5

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

\(\Leftrightarrow15x-3+8x-4=18x\)

\(\Leftrightarrow5x=7\)

hay \(x=\dfrac{7}{5}\)