\(\frac{x+32}{11}=\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\Rightarrow\frac{x+32}{11}+\frac{x+23}{12}-\frac{x+38}{13}-\frac{x+27}{14}=0\)

\(\Rightarrow\left(\frac{x+32}{11}-\frac{x+38}{13}\right)+\left(\frac{x+23}{12}-\frac{x+27}{14}\right)=0\)

\(\Rightarrow\frac{2x-2}{11.13}+\frac{2x-2}{12.14}=0\)

\(\Rightarrow\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)

vì \(\left(\frac{1}{11.13}+\frac{1}{12.14}\right)\ne0\)

mà \(\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)

=> \(\frac{2x-2}{1}=0\Rightarrow2x-2=0\Rightarrow2x=2\Rightarrow x=1\)

bạn tìm ra x chưa

cu nhan cheo bn nha

\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\Rightarrow\)\(\frac{x+32}{11}-3+\frac{x+23}{12}-2=\frac{x+38}{13}-3+\frac{x+27}{14}-2\)

\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)

\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}-\frac{x-1}{14}=0\)

\(\Rightarrow\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\Rightarrow x-1=0\)(Vì \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\))

\(\Rightarrow x=1\)

Vậy:x=1

\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)

\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)

\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)

\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)

\(\Rightarrow x-21=0\Rightarrow x=21\)

\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)

\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)( trừ cả hai vế cho 2 )

\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)

\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}-\frac{x-21}{13}-\frac{x-21}{14}=0\)

\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà  \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

\(\Rightarrow x-21=0\)

\(\Leftrightarrow x=21\)

Vậy  \(x=21\)

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

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

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

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

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

\(3x\left(x-1\right)+5\left(2-x\right)=3x^2-7x+6\) \(6\)

<=> \(3x^2-3x+10-5x=3x^2-7x+6\)

<=> \(-x=-4\)

<=> \(x=4\)

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

<=> \(\left(x+2\right)^2=\frac{1}{6}\)

<=> \(\hept{\begin{cases}x+2=\sqrt{\frac{1}{6}}\\x+2=-\sqrt{\frac{1}{6}}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\sqrt{\frac{1}{6}}-2\\x=-\sqrt{\frac{1}{6}}-2\end{cases}}\)