Question

bài 1 tìm x
a)\(\sqrt{10x+20}\) 
b)\(\sqrt{\dfrac{99}{4x-1}}\)
c)\(\sqrt{\dfrac{-5x-10}{2021}}\)

Answer 1

a)
\(=\sqrt{10\left(x+2\right)}=\sqrt{10}.\sqrt{x+2}\)
\(\sqrt{x+2}\ge0=>x+2\ge0=>x\ge-2\)
b)
\(=\dfrac{\sqrt{99}}{\sqrt{4x-1}}\)
\(=>4x-1>0=>4x>1=>x>\dfrac{1}{4}\)
c)
\(=\dfrac{\sqrt{-5x-10}}{\sqrt{2021}}=>-5x-10\ge0\)
\(=>-5\left(x+2\right)\ge0\)
\(=>x+2\ge0=>x\ge-2\)

Answer 2

a: ĐKXĐ: \(x\ge-2\)

b: ĐKXĐ: \(x>\dfrac{1}{4}\)

c: ĐKXĐ: \(x\le-2\)