Question

cho biểu thức P=\(\frac{x-\sqrt{x}}{x-9}+\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)với x≥0 và x≠9

a)rút gọn P

b)tính giá trị của Ptrong các trường hợp:

i)x=\(\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)

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

Answer 1

a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

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

b/ i, \(x=\sqrt{4+4\sqrt{2}+2}+\sqrt{4-4\sqrt{2}+2}\)

\(=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=2+\sqrt{2}+2-\sqrt{2}=4\)

Thay vào P có:\(P=\frac{2+\sqrt{4}}{3+\sqrt{4}}=\frac{4}{5}\)

ii, \(x=\frac{\sqrt{2}+1-\sqrt{2}+1}{2-1}=2\)

Thay vào có:\(P=\frac{2+\sqrt{2}}{3+\sqrt{2}}=\frac{4+\sqrt{2}}{7}\)

Answer 2

a) Ta có:

\(P=\frac{x-\sqrt{x}}{x-9}+\frac{\sqrt{x}-3}{x-9}-\frac{\sqrt{x}+3}{x-9}\)

\(P=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{x-9}\)

\(P=\frac{x-\sqrt{x}-6}{x-9}\)