a) \(2x=\sqrt{x}\left(ĐK:x\ne0\right)\)
\(\Leftrightarrow\left(2x\right)^2=x\)
\(\Leftrightarrow4.x^2=x\)
\(\Leftrightarrow4=x:x^2=x.\frac{1}{x^2}=\frac{x}{x^2}=x^{-1}\) ( vô lí vì \(x^{-1}\le0\) )
Vậy : \(x\in\varnothing\)
b) \(\sqrt{x}-1=2\)
\(\Leftrightarrow\sqrt{x}=2+1=3\)
\(\Leftrightarrow x=3^2=9\)
Vậy : \(x=9\)
c) \(3\sqrt{x}-2=7\)
\(\Leftrightarrow3\sqrt{x}-2=7+2=9\)
\(\Leftrightarrow\sqrt{x}=9:3=3\)
\(\Leftrightarrow x=3^2=9\)
Vậy :\(x=9\)
d) \(\sqrt{x-1}+1=3\)
\(\Leftrightarrow\sqrt{x-1}=3-1=2\)
\(\Leftrightarrow x-1=2^2=4\)
\(\Leftrightarrow x=5\)
Vậy : \(x=5\)
b) \(\sqrt{x}-1=2\)
=> \(\sqrt{x}=2+1\)
=> \(\sqrt{x}=3\)
=> \(3^2=9\)
=> \(x=9\)
Vậy \(x=9.\)
c) \(3\sqrt{x}-2=7\)
=> \(3\sqrt{x}=7+2\)
=> \(3\sqrt{x}=9\)
=> \(\sqrt{x}=9:3\)
=> \(\sqrt{x}=3\)
=> \(3^2=9\)
=> \(x=9\)
Vậy \(x=9.\)
d) \(\sqrt{x-1}+1=3\)
=> \(\sqrt{x-1}=3-1\)
=> \(\sqrt{x-1}=2\)
=> \(x-1=2^2\)
=> \(x-1=4\)
=> \(x=4+1\)
=> \(x=5\)
Vậy \(x=5.\)
Chúc bạn học tốt!
