Question

\(\left(2x-1\right)^{10}=49^5\)

Answer 1

`=> (2x-1)^2 = 49`

`=>` \(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Answer 2

\(\left(2x-1\right)^{10}=49^5=\left(7^2\right)^5=7^{2.5}\\ =>\left(2x-1\right)^{10}=7^{10}=\left(\pm7\right)^{10}\\ =>\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Answer 3

`(2x-1)^10 = 49^5`

`=>` \(\left(2x-1\right)^{10}=7^{2.5}\)

`=>` \(\left[{}\begin{matrix}\left(2x-1\right)^{10}=7^{10}\\\left(2x-1\right)^{10}=\left(-7\right)^{10}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Answer 4

\(\left(2x-1\right)^{10}=49^5\)

\(\left(2x-1\right)^{10}=\left(7^2\right)^5\)

\(\left(2x-1\right)^{10}=7^{10}\)

\(\Rightarrow2x-1=7\)

\(2x=7+1\)

\(2x=8\)

\(x=\dfrac{8}{2}\)

\(x=4\)

Vậy \(x=4\)