\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)

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

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

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

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

\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)

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

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

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

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

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

b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep

c, tt

d, cx r

a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)

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

b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)

bạn không ghi yêu cầu nên mình làm như này

1) \(\frac{1}{x-3}\) và \(\frac{5}{x^2-3x}\)

Ta có: \(1.\left(x^2-3x\right)=x^2-3x\)

           \(\left(x-3\right).5=5x-15\)

\(\Rightarrow x^2-3x\ne5x-15\)

\(\Rightarrow1.\left(x^2-3x\right)\ne\left(x-3\right).5\)

Vậy: \(\frac{1}{x-3}\ne\frac{5}{x^2-3x}\)

2) \(\frac{x}{x^2+x}\) và \(\frac{2}{x-1}\) và \(\frac{x+2}{x^2-1}\)

Ta có: \(x.\left(x-1\right)=x^2-x\)

          \(2.\left(x^2+x\right)=2x^2+2x\)

\(\Rightarrow x^2-x\ne2x^2+2x\)

\(\Rightarrow x.\left(x-1\right)\ne2.\left(x^2+x\right)\)

\(\Rightarrow\frac{1-3x}{2x}\ne\frac{2}{x-1}\) (1)

Ta lại có: \(2.\left(x^2-1\right)=2x^2-2\)

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

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

\(\Rightarrow2x^2-2\ne x^2-x-2\)

\(\Rightarrow2.\left(x^2-1\right)\ne\left(x-1\right)\left(x+2\right)\)

\(\Rightarrow\frac{2}{x-1}\ne\frac{x+2}{x^2-1}\) (2)

Từ (1) và (2) => \(\frac{x}{x^2+x}\ne\frac{2}{x-1}\ne\frac{x+2}{x^2-1}\)

3) \(\frac{1-3x}{2x}\) và \(\frac{3x-2}{2x-1}\) và \(\frac{3x-2}{4x^2-2x}\)

Ta có:\(\left(1-3x\right)\left(2x-1\right)=2x-1-6x^2+3x\)

                                                   \(=5x-1-6x^2\)

          \(2x.\left(3x-2\right)=6x^2-4x\)

\(\Rightarrow5x-1-6x^2\ne6x^2-4x\)

\(\Rightarrow\left(1-3x\right)\left(2x-1\right)\ne2x\left(3x-2\right)\)

\(\Rightarrow\frac{1-3x}{2x}\ne\frac{3x-2}{2x-1}\)(1)

Ta lại có: \(\left(3x-2\right)\left(4x^2-2x\right)=12x^2-6x^2-8x^2+4x\)

                                                             \(=12x^3-14x^2+4x\)

                \(\left(2x-1\right)\left(3x-2\right)=6x^2-4x-3x+2\)

                                                         \(=6x^2-7x+2\)

\(\Rightarrow12x^3-14x^2+4x\ne6x^2-7x+2\)

\(\Rightarrow\left(3x-2\right)\left(4x^2-2x\right)\ne\left(2x-1\right)\left(3x-2\right)\)

\(\Rightarrow\frac{3x-2}{2x-1}\ne\frac{3x-2}{4x^2-2x}\) (2)

Từ (1) và (2) => \(\frac{1-3x}{2x}\ne\frac{3x-2}{2x-1}\ne\frac{3x-2}{4x^2-2x}\)

\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)

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

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

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

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

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

a.ĐK: 2x²+1\(\ne0\) \(\forall x\)

Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x²+1 >0 với mọi x)

\(\Leftrightarrow x=2\) (TM)

Vậy ...

b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)

Để phương trình bằng 0 thì x²-x-6=0 (Vì x-3\(\ne0\))

\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)

Vậy ...

c. ĐK: x\(\ne\)2

\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)

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

\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)

\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)

Vậy ...

d. ĐK: \(x\ne\pm\frac{1}{3}\)

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)

\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))

\(\Leftrightarrow x=-2\:\left(TM\right)\)

Vậy...

Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)

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

ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)

\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)

\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)

Thỏa mãn ĐK

Các trường hợp khác làm tương tự

a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x²
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :

\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)

\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)

Đến đây ta đặt  \(x+\frac{60}{x}+16=t\left(1\right)\)

Ta được :

\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)

Từ đó ta lắp vào ( 1 ) tính được x 

a) Ta có: \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)

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

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

\(=\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)

\(=\frac{x^2+5x+6}{2x\left(x+3\right)}\)

\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}\)

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

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

b) Ta có: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)

\(=\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)

\(=\frac{3x}{x\left(2x+6\right)}-\frac{x-6}{x\left(2x+6\right)}\)

\(=\frac{3x-x+6}{x\left(2x+6\right)}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)

c) Ta có: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)

\(=\frac{5\left(x+2\right)\cdot2\cdot\left(2-x\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}\)

\(=\frac{5\cdot2\cdot\left(2-x\right)}{-4\left(2-x\right)}=\frac{5\cdot2}{-4}=\frac{-5}{2}\)

d) Ta có: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)

\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3x}{x\left(x+4\right)\cdot2\left(2-x\right)}\)

\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3}{2\left(x+4\right)\cdot\left(2-x\right)}=\frac{3\left(1-4x^2\right)}{2\left(-x^2-2x+8\right)}\)

\(=\frac{3-12x^2}{-2x^2-4x+16}\)

a) \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)

\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne-3;x\ne0\right)\)

\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{4x+6}{2x\left(x+3\right)}\)

\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)

b) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne0;x\ne-3\right)\)

\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)

c) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}\) \(\left(ĐKXĐ:x\ne\pm2\right)\)

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

a)\(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

   \(84x+63-90x+30=175x+140+315\)

    93-6x=175x+455

     -362=181x

       x=-2

b)\(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)

   \(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)

      \(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)

        \(\left(3x+1\right)\left(-x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)