Xét bài toán phụ sau:

Nếu \(a+b+c=0\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)  \(\left(a,b,c\ne0\right)\)

Thật vậy

Ta có: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{a+b+c}{abc}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{0}{abc}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

Bài toán được chứng minh

Quay trở lại, ta sẽ áp dụng bài toán phụ vào bài chính:

Ta có: \(P=\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}+...+\sqrt{\frac{1}{2^2}+\frac{1}{779^2}+\frac{1}{801^2}}\)

Vì \(2+1+\left(-3\right)=0\) nên:

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

Tương tự ta tính được:

\(\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}=\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\) ; ... ; \(\sqrt{\frac{1}{2^2}+\frac{1}{799^2}+\frac{1}{801^2}}=\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)

\(\Rightarrow P=\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)

\(=\frac{1}{2}\cdot400+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{799}-\frac{1}{801}\right)\)

\(=200+\frac{800}{801}=\frac{161000}{801}=\frac{a}{b}\Rightarrow\hept{\begin{cases}a=161000\\b=801\end{cases}}\)

\(\Rightarrow Q=161000-801\cdot200=800\)

Với mọi n thuộc N ta có :

\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+\frac{2}{n}-\frac{2}{n\left(n+1\right)}-\frac{2}{\left(n+1\right)}}\)

\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}=1+\frac{1}{n}-\frac{1}{n+1}\)

Áp dụng ta được :

\(S=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+....+\left(1+\frac{1}{99}-\frac{1}{100}\right)\)

\(=98+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)

Bài 1:

a, \(4\sqrt{3+2\sqrt{2}}-\sqrt{57+40\sqrt{2}}\)

\(=4\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(4\sqrt{2}+5\right)^2}\)

\(=4\left(\sqrt{2}+1\right)-4\sqrt{2}-5\)

\(=4\sqrt{2}+4-4\sqrt{2}-5=-1\)

b, \(B=\sqrt{1100}-7\sqrt{44}+2\sqrt{176}-\sqrt{1331}\)

\(=10\sqrt{11}-14\sqrt{11}+8\sqrt{11}-11\sqrt{11}=-7\sqrt{11}\)

c, \(C=\sqrt{\left(1-\sqrt{2002}\right)^2}.\sqrt{2003+2\sqrt{2002}}\)

\(=\left(1-\sqrt{2002}\right).\sqrt{\left(\sqrt{2002}+1\right)^2}\)

\(=\left(1-\sqrt{2002}\right).\left(\sqrt{2002}+1\right)=-2001\)

Câu d bạn kiểm tra lại đề bài nhé.

Bài 2:

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

a, ĐK: \(x\ge0,x\ne1\)

b, ĐK: \(x\ge0,x\ne1\)

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

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

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

\(=\frac{2\sqrt{x}+2-2\sqrt{x}+2}{4\left(x-1\right)}-\frac{\sqrt{x}}{x-1}\)

\(=\frac{4-4\sqrt{x}}{4\left(x-1\right)}=\frac{4\left(1-\sqrt{x}\right)}{4\left(1-x\right)}=\frac{1-\sqrt{x}}{1-x}\)

Thay \(x=3\left(TM\right)\)vào A ta có: \(A=\frac{1-\sqrt{3}}{3-1}=\frac{1-\sqrt{3}}{2}\)

Vậy với \(x=3\)thì \(A=\frac{1-\sqrt{3}}{2}\)

c, \(\left|A\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}A=\frac{1}{2}\\A=-\frac{1}{2}\end{cases}}\)

TH1: \(A=\frac{1}{2}\)\(\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=\frac{1}{2}\Leftrightarrow2-2\sqrt{x}=x-1\)\(\Leftrightarrow x-1-2+2\sqrt{x}=0\)\(\Leftrightarrow x+2\sqrt{x}-3=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\\sqrt{x}=-3\left(L\right)\end{cases}}}\)

TH2: \(A=-\frac{1}{2}\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=-\frac{1}{2}\)\(\Leftrightarrow2-2\sqrt{x}=1-x\Leftrightarrow-x+1-2+2\sqrt{x}=0\)\(\Leftrightarrow-x-1+2\sqrt{x}=0\Leftrightarrow x-2\sqrt{x}+1=0\)\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\Leftrightarrow\sqrt{x}=-1\left(L\right)\)

Vậy với \(x=1\)thì \(\left|A\right|=\frac{1}{2}\)

Cám ơn bạn nhiều nha!!!

Với \(k\in N;k\ne0\) ta có :

\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{\left(k+1\right)}}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}\)

\(=\frac{\sqrt{k+1}+\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\)

\(=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

Áp dụng ta có :

\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)

\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)