giúp mik vs

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

<=>\(\frac{2}{3}x+x=\frac{4}{3}+\frac{3}{4}\)

<=>\(\frac{5}{3}x=\frac{25}{12}\)

<=>\(x=\frac{5}{4}\)

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

\(\frac{2}{3}x+x=\frac{4}{3}+\frac{3}{4}\)

\(\frac{5}{3}x=\frac{25}{12}\)

      \(x=\frac{25}{12}:\frac{5}{2}\)

      \(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\)

c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)

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

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

Từ (1) và (2) => 2x+3y-z=\(2x+3\left(\frac{3\left(x-1\right)+4}{2}\right)-\frac{4\left(x-1\right)+6}{2}=50\)

\(\Rightarrow\frac{4x}{2}+\frac{9\left(x-1\right)+12}{2}-\frac{4\left(x-1\right)+6}{2}=50\)

\(\Rightarrow\frac{4x+9x-9+12-4x+4-6}{2}=50\)

\(\Rightarrow9x+1=100\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=11\)

Vì \(y=\frac{3\left(x-1\right)+4}{2}=\frac{3\left(11-1\right)+4}{2}=\frac{34}{2}=17\Leftrightarrow y=17\)

Vì \(z=\frac{4\left(x-1\right)+6}{2}=\frac{4\left(11-1\right)+6}{2}+\frac{46}{2}=23\Leftrightarrow z=23\)

Vậy   x=11

         y=17

         z=23

\(\Rightarrow\frac{2\left(x-1\right)}{2.2}=\frac{3\left(y-2\right)}{3.3}=\frac{z-3}{4}\) 

\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\) 

Áp dụng t/c dãy tỉ số = nhau

\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{50-2-6+3}{9}=\frac{45}{9}=5\) 

\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2}=5\Rightarrow x-1=10\Rightarrow x=11\\\frac{y-2}{3}=5\Rightarrow y-2=15\Rightarrow y=17\\\frac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow z=23\end{cases}}\)

Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)=> \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{-z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{-z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}\)= \(\frac{50-2-6+3}{9}=\frac{45}{9}=5\).

=> \(\hept{\begin{cases}\frac{x-1}{2}=5\\\frac{y-2}{3}=5\\\frac{z-3}{4}=5\end{cases}}\)=> \(\hept{\begin{cases}x-1=10\\y-2=15\\z-3=20\end{cases}}\)=> \(\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}\).

a) \(\frac{x^2+5x}{5x^2+x^3}\)

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

b) \(\frac{x^4+x^2+1}{x^3+1}\)

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

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

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

\(a,\frac{x^2+5x}{5x^2+x^3}=\frac{x\left(x+5\right)}{x^2\left(x+5\right)}=\frac{1}{x}\)

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

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

hok tốt nha

\(-x-\frac{3}{4}=-\frac{8}{11}=>-x=-\frac{8}{11}+\frac{3}{4}=\frac{1}{44}=>x=-\frac{1}{44}\)

-x-3/4=-8/11

=0.02272727....

11/12-(2/5+x)=2/3

=-0.15

- x - 34 = −811    

 -x         = \(\frac{-8}{11}+\frac{3}{4}\)        

-x         = \(\frac{-32}{44}+\frac{33}{44}\)   

 -x       = \(\frac{1}{44}\)

=> x   =  \(\frac{-1}{44}\) 

\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

               \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}\) 

               \(\frac{2}{5}+x=\frac{11}{12}-\frac{8}{12}\) 

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

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

                            \(x=\frac{5}{20}-\frac{-8}{20}\) 

                          \(\Rightarrow x=\frac{-3}{20}\)

Ủng hộ nha!^^