Question

tìm x biết

a.\(\sqrt{x-2}=12\)

b.\(\sqrt{x-1}=\frac{1}{3}\)

c.\(\sqrt{2x+\frac{5}{4}}=\frac{3}{2}\)

nhanh lên nha, 9h hoặc 9h30 mk đi hok rồi

giúp nha

Answer 1

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 \(x=\frac{1}{2}\)

Answer 2

Answer 3

a.\(\sqrt{x-2}\) =12 <=> x-2 = 12\(^2\) =144 <=>x= 146

b.\(\sqrt{x-1}\) = \(\frac{1}{3}\) <=>x-1= (\(\frac{1}{3}\) )\(^2\) =\(\frac{1}{9}\) <=> x= \(\frac{10}{9}\)

c.\(\sqrt{2x+\frac{5}{4}}\) =\(\frac{3}{2}\) <=> 2x+\(\frac{5}{4}\) = (\(\frac{3}{2}\))\(^2\) = \(\frac{9}{4}\) <=> 2x= 1 <=> x=\(\frac{1}{2}\)

Answer 4

thanks 2 bn nha