Question

Tìm \(x\) không âm biết :

a) \(\sqrt{x}=3\)

b) \(\sqrt{x}=\sqrt{5}\)

c) \(\sqrt{x}=0\)

d) \(\sqrt{x}=-2\)

Answer 1

\(\sqrt{9}=3\)

\(\sqrt{25=3}\)

\(\sqrt{0}=0\)

\(-\sqrt{4}\)

Answer 2

a, \(\sqrt{x}\)=3 ( đkxđ : \(x\ge0\))

<=> \(\left(\sqrt{x}\right)^{^{ }2}\)= \(^{3^2}\)

<=> x = 9

b, \(\sqrt{x}\)= \(\sqrt{5}\) ( đkxđ : \(x\ge0\))

<=> \(\left(\sqrt{x}\right)^2=\left(\sqrt{5}\right)^2\)

<=> x = 5

c, \(\sqrt{x}=0\) ( đkxđ : \(x\ge0\))

<=> \(\left(\sqrt{x}\right)^2=0^2\)

<=> x = 0

d, \(\sqrt{x}=-2\) ( đkxđ : \(x\ge0\))

vô nghiệm

Vậy k có giá trị nào của x ( tm đkxđ)

Answer 3

a) \(\sqrt{x}\)=3 ( ĐKXĐ: x\(\ge\)3)

<=> (\(\sqrt{x}\))²=3²

<=>x=9

Vậy x=9

b)\(\sqrt{x}=\sqrt{5}\) ( ĐKXĐ : x\(\ge\)0)

<=>\(\left(\sqrt{x}\right)^2=\left(\sqrt{5}\right)^2\)

<=> x=5

Vậy x=5

c) \(\sqrt{x}=0\) (ĐKXĐ : x\(\ge\)0)

<=> \(\left(\sqrt{x}\right)^2=0^2\)

<=> x=0

Vậy x=0

d) \(\sqrt{x}=-2\) ( ĐKXĐ : x\(\ge\)0 )

<=>\(\left(\sqrt{x}\right)^2=\left(-2\right)^2\)

<=> x=4

Vậy x=4