\(\sqrt{9}=3\)

\(\sqrt{25=3}\)

\(\sqrt{0}=0\)

\(-\sqrt{4}\)

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)

Với câu c, Thiên Anh nên thêm điều kiện để phần kết luận là: \(0\le x< 2.\)

xin lỗi 

mình không làm được

a, 67/57

b,Q =678/78 n/t

c, s = a+h

Ta có : \(\sqrt{3}.x-\sqrt{75}=0\)

\(\Leftrightarrow\sqrt{3}.x-5\sqrt{3}=0\)

\(\Leftrightarrow\sqrt{3}\left(x-5\right)=0\)

Vì \(\sqrt{3}\ne0\)

Nên : x - 5 = 0

Vậy x = 5. 

b) Ta có : \(\sqrt{2}.x+\sqrt{2}=\sqrt{8}+\sqrt{32}\)

\(\Leftrightarrow\sqrt{2}\left(x+1\right)=6\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\left(x+1\right)-6\sqrt{2}=0\)

\(\Leftrightarrow\sqrt{2}.\left(x+1-6\right)=0\)

\(\Leftrightarrow\sqrt{2}.\left(x-5\right)=0\)

Vì \(\sqrt{2}\ne0\)

Nên x - 5 = 0

Suy ra : x = 5

Bài 1:

a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)

= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)

=\(\sqrt{xy}\)

b.ĐK: x ≠ 1

Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)

*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)

⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)

⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)

c.Ta có:

a) \(\sqrt{2}\cdot x-\sqrt{50}=0< =>\sqrt{2}\cdot x=\sqrt{50}\)

<=> x= 5

b) \(\sqrt{3}\cdot x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)

<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot\sqrt{4}+\sqrt{3}\cdot\sqrt{9}\)

<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot5< =>x+1=5\)

<=> x=4

c) \(\sqrt{3}\cdot x^2-\sqrt{12}=0\\ < =>x^2=\sqrt{4}=2;-2\\ < =>x=\sqrt{2};-\sqrt{2}\)

d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\\ < =>x^2=\sqrt{100}=10;-10\\ < =>x=\sqrt{10};-\sqrt{10}\)