a) \(\sqrt{x-2}=12\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow x-2=144\)

\(\Leftrightarrow x=146\) (tm)

Vậy x=146

b)\(\sqrt{x-1}=\frac{1}{3}\left(ĐK:x\ge1\right)\)

\(\Leftrightarrow x-1=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{10}{9}\left(tm\right)\)

Vậy x=\(\frac{10}{9}\)

c)\(\sqrt{2x+\frac{5}{4}}=\frac{3}{2}\left(ĐK:x\ge\frac{-5}{8}\right)\)

\(\Leftrightarrow2x+\frac{5}{4}=\frac{9}{4}\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)

vậy \(x=\frac{1}{2}\)

MÌNH CẦN GẤP NHA MN

\(x^2=16\)

=>x=4

b)

=>(x+2)^2=7^2

x+2=7

x=5

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=\pm\sqrt{\frac12}\end{array}\right.\)

Vậy \(x\) ∈ {- \(\sqrt{\frac12}\); 0; \(\sqrt{\frac12}\)}

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=-\frac{1}{\sqrt2}\\ x=\frac{1}{\sqrt2}\end{array}\right.\)

Vậy \(x\) \(\in\) {- \(\frac{1}{\sqrt2}\); 0; \(\frac{1}{\sqrt2}\)}

b: =>căn x+1=-2(loại)

c: =>|x+1|=3

=>x+1=3 hoặc x+1=-3

=>x=-4 hoặc x=2

d: =>x-3=16

=>x=19