Dạng tổng quát :
\(\frac{1}{\sqrt{a-1}+\sqrt{a}}=\frac{\sqrt{a-1}-\sqrt{a}}{\left(\sqrt{a-1}+\sqrt{a}\right)\left(\sqrt{a-1}-\sqrt{a}\right)}\)
\(=\frac{\sqrt{a-1}-\sqrt{a}}{a-1-a}=\sqrt{a}-\sqrt{a-1}\)
Do đó :
\(VT=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-\sqrt{1}\)
\(=10-1=9\)
Chứng minh:
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=9\)
\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+2}+...+\frac{1}{\sqrt{100}+\sqrt{99}}=9\)
Tính tổng sau:
S=\(\frac{1}{2\sqrt[]{1}+1\sqrt[]{2}}+\frac{1}{3\sqrt[]{2}+2\sqrt[]{3}}+.........+\frac{1}{100\sqrt[]{99}+99\sqrt[]{100}}\)
Tính tổng : T=\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{99}-\sqrt{100}}\)
\(\frac{1+\sqrt{1}+\sqrt{2}}{1-\sqrt{1}+\sqrt{2}}+...+\frac{1+\sqrt{99}+\sqrt{100}}{1-\sqrt{99}+\sqrt{100}}\)
\(\frac{1-\sqrt{1}+\sqrt{2}}{1+\sqrt{1}+\sqrt{2}}+...+\frac{1-\sqrt{99}+\sqrt{100}}{1+\sqrt{99}+\sqrt{100}}\)
Tính:
a) \(A=\frac{1-ax}{1+ax}\sqrt{\frac{1+bx}{1-bx}}\) tại \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\)
b) \(B=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
c) \(C=\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}\)
Bài 1: Tính :
\(C=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(D=\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Bài 2 : Cho \(P=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{x-\sqrt{x}-2}{x+\sqrt{x}+2}\right)\)
a, Rút gọn P
b, Tìm GTNN
c, Tìm x để \(P.\frac{x-1}{x^2+8x}< -2\)
C/Minh đẳng thức:
a) \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{a-1}\) (với a>0, b>0, a≠b)
b)\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\) (với a>0, b>0,a≠b)
c) \(\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}=\frac{a+9}{a-9}\) (với a≥0, b≥0,a≠9)
