cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11) 
 

TH1: x-7=0 => x=7 => 0^8=0^18 (TM) 
 

TH2: x-7=1 => x=8 (TM) 
 

TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại 
 

KL: x = 7 hoặc x=8

( x-7)^( x+1) - ( x-7)^(x+11) = 0 
 

( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0 
 

( x-7)^( x+1) (1-x^10) = 0 

tới đây dễ òi

a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)

<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

<=> x = 2010

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

Ta thấy : \(\left|x-1\right|\ge0;\left|x-2\right|\ge0;\left|x-3\right|\ge0\)

=> \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge0\)

=> 4 ( x - 4 ) \(\ge0\). Mà 4 > 0 => \(x-4\ge0=>x\ge4\)hay

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)=>x-1+x-2+x-3=4\left(x-4\right)\) => 3x - 6 = 4x - 16

=> -6+16 = 4x - 3x => x = 10

kfckfckfc ngon ngonngon

tich mik minh tich lai

x=10 cách lm thì chờ xem lại

\(\left(\frac{2}{7}\right)^{6x-7}=1\)

\(\Rightarrow\left(\frac{2}{7}\right)^{6x-7}=\left(\frac{2}{7}\right)^0\)

\(\Rightarrow6x-7=0\)

\(\Rightarrow6x=7\Leftrightarrow x=\frac{7}{6}\)