\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
\(\Rightarrow\left(2x+1\right)^6-\left(2x+1\right)^4=0\)
\(\Rightarrow\left(2x+1\right)^4.\left[\left(2x+1\right)^2-1\right]=0\)
\(\Rightarrow\left(2x+1\right)^4=0\) hoặc \(\left(2x+1\right)^2-1=0\)
+) \(\left(2x+1\right)^4=0\Rightarrow2x+1=0\Rightarrow x=-0,5\)
+) \(\left(2x+1\right)^2-1=0\Rightarrow\left(2x+1\right)^2=1\)
\(\Rightarrow2x+1=\pm1\)
+ \(2x+1=1\Rightarrow x=0\)
+ \(2x+1=-1\Rightarrow x=-1\)
Vậy \(x\in\left\{-0,5;0;-1\right\}\)
(2x+1)4=(2x+1)6
\(\Leftrightarrow\)16x+1=64x+1
\(\Leftrightarrow\)16x-64x=1-1
\(\Leftrightarrow\)-48x=0
\(\Leftrightarrow\)x=0
mik ko chắc..
