\(\left[2.\left(x+\frac{1}{2}\right)\right]^4=\frac{1}{16}\)
\(\left[2.\left(x+\frac{1}{2}\right)\right]^4=\left(\frac{1}{2}\right)^4\)
\(2.\left(x+\frac{1}{2}\right)=\frac{1}{2}\)
\(x+\frac{1}{2}=\frac{1}{2}.\frac{1}{2}\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=\frac{1}{4}-\frac{2}{4}\)
\(x=\frac{-1}{4}\)
\(2.\left(x+\frac{1}{2}\right)^4=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^4=\frac{1}{16}:2\)
\(\left(x+\frac{1}{2}\right)^4=\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
...
-_-
mk ko bk lm và vs đề bài bn ghi như z, mk nghĩ sẽ ko tìm đk x đâu!
