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

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1 tháng 12 2019

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

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

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

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

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

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

14 tháng 12 2018

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

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

14 tháng 12 2018

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

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

13 tháng 7 2017

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

25 tháng 4 2017

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

25 tháng 4 2017

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

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

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

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

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

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

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

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

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

27 tháng 3 2020

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

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

21 tháng 5 2021

\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)

\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)

\(< =>12x-20-14x-21=0\)

\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)

21 tháng 5 2021

\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)

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

\(< =>8x+12+4x-2x+3=0\)

\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)

16 tháng 12 2018

\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)

\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)

\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)

\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)

\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)

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

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

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

5 tháng 12 2018

1/ \(\frac{x-3}{3xy}\)+\(\frac{5x+3}{3xy}\)\(\frac{6x}{3xy}\)=\(\frac{3}{y}\)

2/\(\frac{5x-7}{2x-3}\)+\(\frac{4-3x}{2x-3}\)=\(\frac{2x-3}{2x-3}\)=1

3/\(\frac{11x-7}{3-5x}\)-\(\frac{6x+4}{5x-3}\)=\(\frac{11x-7}{3-5x}\)+\(\frac{6x+4}{3-5x}\)=\(\frac{17x-3}{3-5x}\)

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

5/\(\frac{1}{2x-10}\)+\(\frac{2x}{3x^2-15x}\)=\(\frac{1}{2\left(x-5\right)}\)+\(\frac{2x}{3x\left(x-5\right)}\)=\(\frac{3x}{6x \left(x-5\right)}\)+\(\frac{4x}{6x\left(x-5\right)}\)

=\(\frac{7x}{6x\left(x-5\right)}\)=\(\frac{7}{6\left(x-5\right)}\)