Question

tính giá trị biểu thức:

\(\frac{x-9}{\sqrt{x}+3}+\frac{2\sqrt{x}-6}{\sqrt{x}-3}\)   (x>0,#+-9)

Tìm x biết 

a.\(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\)

b.\(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

giúp mình voi

 

Answer 1

Bài 1:

\(\frac{x-9}{\sqrt{x}+3}+\frac{2\sqrt{x}-6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

\(=\sqrt{x}-3+2=\sqrt{x}-1\)

Bài 2:

a) Không rõ đề

b) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{3}+1\\x-3=-\sqrt{3}-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\)