Mình thiếu điều kiện xác định ^_^

Cho mình bổ xung thêm

\(ĐKXĐ:x\ne\pm1\)

và mình sửa lại nữa là: \(\orbr{\begin{cases}x=-1\left(L\right)\\x=-3\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{-3\right\}\)

\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{x^2+3}{1-x^2}\) đkxđ \(x\ne\pm1\)

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

\(\Leftrightarrow x^2+2x+1-x^2-2x-1+x^2+3=0\)

\(\Leftrightarrow x^2+3=0\)

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

\(\Leftrightarrow x\in\varnothing\)

\(ĐKXĐ:x\ne\pm1\)

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

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=-x^2-3\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=-x^2-3\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-1\end{cases}}\)

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

So sánh với ĐKXĐ ta thấy: \(x=0\)không thoả mãn

Vậy tập nghiệm của phương trình là \(S=\left\{-1\right\}\)

Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

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

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

    \(\Rightarrow x^2+x+2=2\)

   \(\Leftrightarrow x^2+x=0\)

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

   \(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy \(S=\left\{-1;0\right\}\)

\(ĐKXĐ:x\ne2;x\ne0\)

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

\(\Leftrightarrow x^2+2x-x+1=2\)

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

\(\Delta=1^2+4.1=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+\sqrt{5}}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{cases}}\)

\(ĐKXĐ:x\ne-1;x\ne\frac{2}{3}\)

\(pt\Leftrightarrow\frac{7x-2\left(x+1\right)+\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=1\)

\(\Leftrightarrow7x-2\left(x+1\right)+\left(3x-2\right)=\left(3x-2\right)\left(x+1\right)\)

\(\Leftrightarrow8x-4=3x^2-2x+3x-2\)

\(\Leftrightarrow3x^2-7x+2=0\)

\(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+5}{6}=2\\x=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)

Tự cho đkxđ nha!!!

<=> \(\frac{x+1-x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2}{3x-2}\)

<=> \(\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(3x-2\right)\left(x+1\right)}\)

<=> \(\frac{7x-2x-2-3x+2}{\left(3x-2\right)\left(x+1\right)}=0\)

<=> \(\frac{2x}{\left(3x-2\right)\left(x+1\right)}=0\)

=> 2x = 0

<=> x = 0 (TM)

Vậy ...

\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\)

\(\left(x+1\right)\left(3x-2\right)\left(2-3x\right)-x\left(3x-2\right)\left(2-3x\right)=7x\left(2-3x\right)+2\left(x+1\right)\left(3x-2\right)\)

\(-9x^2+12x-4=16x-15x^2-4\)

\(-9x^2+12x=16x-15x^2\)

\(9x^2-12x+16x-15x^2=0\)

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

\(-2x\left(3x-2\right)=0\)

\(Th1:-2x=0\Leftrightarrow x=0\)

\(Th2:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

vô nghiệm

Ta có : \(\frac{1+3y}{12}=\frac{1+6y}{16}\)

<=> (1 + 3y).16 = (1 + 6y).12

<=> 16 + 48y = 12 + 72y

<=> 16 - 12 = 72y - 48y

<=> 24y = 4

=> y = 1/6 

Thay y = 1/6 vào ta có : \(\frac{1+6.\frac{1}{6}}{16}=\frac{1+9.\frac{1}{6}}{4x}\Rightarrow\frac{1}{8}=\frac{\frac{5}{2}}{4x}\) 

=> x = \(\frac{5}{2}:\frac{1}{8}=20\)

Đkxđ: \(\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)

\(\frac{x+3}{x-2}+\frac{x+2}{x}=2\) 

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

\(\Rightarrow x\left(x+3\right)+\left(x-2\right)\left(x+2\right)=2x\left(x-2\right)\)

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

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

\(\Leftrightarrow7x=4\)

\(\Leftrightarrow x=\frac{4}{7}\)

a.

\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)

TH1:

\(x+\frac{1}{2}=0\)

\(x=-\frac{1}{2}\)

TH2:

\(x-\frac{3}{4}=0\)

\(x=\frac{3}{4}\)

Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)

b.

\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)

TH1:

\(\frac{1}{2}x-3=0\)

\(\frac{1}{2}x=3\)

\(x=3\div\frac{1}{2}\)

\(x=3\times2\)

\(x=6\)

TH2:

\(\frac{2}{3}x+\frac{1}{2}=0\)

\(\frac{2}{3}x=-\frac{1}{2}\)

\(x=-\frac{1}{2}\div\frac{2}{3}\)

\(x=-\frac{1}{2}\times\frac{3}{2}\)

\(x=-\frac{3}{4}\)

Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)

c.

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

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

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

\(-\frac{4}{3}x=\frac{13}{3}\)

\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)

\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)

\(x=-\frac{13}{4}\)

d.

\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)

\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)

\(x=5\)

\(ĐKXĐ:x\ne\pm3\)

\(pt\Leftrightarrow\frac{\left(x+3\right)^2-\left(x-3\right)^2}{x^2-9}=\frac{17}{x^2-9}\)

\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=17\)

Tự dừng bấm Gửi tl

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=17\)

\(\Leftrightarrow12x=17\Leftrightarrow x=\frac{17}{12}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-9}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-3^2}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{\left(x+3\right)\left(x-3\right)}=\frac{x-3}{x+3}\)

\(\left(x+3\right)^2-17=\left(x-3\right)^2\)

\(x^2+6x-8=x^2-6x+9\)

\(6x-8=-6x+9\)

\(6x=-6x+9+8\)

\(6x+6x=17\)

\(12x=17\Leftrightarrow x=\frac{17}{12}\)