\(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left(2x-3\right)^2=6^2\)
\(\Leftrightarrow2x-3=6\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{9}{2}\)
Tìm x biết:
(2x−3)2= 36
=>(2x-3)2=62
=>2x-3=6
2x=9
x=9:2
x=4,5
Vậy x=4,5
\(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)=\sqrt{36}\)
\(\left(2x-3\right)=6\)
\(2x=6+3\)
\(2x=9\)
\(x=9:2\)
\(x=\frac{9}{2}\)
( 2x - 3 )2 = 36
=> ( 2x - 3 )2 = 62
=> 2x - 3 = 6
=> 2x = 6 + 3
=> 2x = 9
=> x = 9 : 2
=> x = 4,5
\(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left(2x-3\right)^2=6^2\)
\(\Leftrightarrow2x-3=6\)
\(\Leftrightarrow2x=6+3\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=9:2\)
\(\Leftrightarrow x=\frac{9}{2}\)
\(\left(2x-3\right)^2=36\)
\(< =>\left(2x-3\right)^2=6^2\)
\(< =>2x-3=6\)
\(< =>2x=6+3\)
\(< =>2x=9\)
\(< =>x=\frac{9}{2}\)
\(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x< 3\Rightarrow2x-3=-6\\2x>3\Rightarrow2x-3=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-3\\2x=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\x=4,5\end{matrix}\right.\)
Vậy \(x=\left\{-1,5;4,5\right\}\)
