Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Mình đã tìm ra cách giải rồi, các bạn có thể góp ý để bài làm của mình hoàn thiện hơn nữa nha...
Ta có:\(\frac{1}{A}=\frac{\sqrt{a-2003}+\sqrt{b-2003}}{\sqrt{a+b}}=\frac{\sqrt{a-2003}}{\sqrt{a+b}}+\frac{\sqrt{b-2003}}{\sqrt{a+b}}\)
Mặt khác:\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2003}\Rightarrow\frac{a+b}{ab}=\frac{1}{2003}\Rightarrow2003=\)\(\frac{ab}{a+b} \left(1\right)\)
Thay (1) vào \(\frac{1}{A}\) ta được: \(\frac{1}{A}=\frac{\sqrt{a-\frac{ab}{a+b}}}{\sqrt{a+b}}+\frac{\sqrt{b-\frac{ab}{a+b}}}{\sqrt{a+b}}\)
\(\Leftrightarrow\frac{1}{A}=\sqrt{\frac{a-\frac{ab}{a+b}}{a+b}}+\sqrt{\frac{b-\frac{ab}{a+b}}{a+b}}\)
\(\Leftrightarrow\frac{1}{A}=\sqrt{\frac{\frac{a^2+ab-ab}{a+b}}{a+b}}+\sqrt{\frac{\frac{b^2+ab-ab}{a+b}}{a+b}}=\sqrt{\frac{a^2}{\left(a+b\right)^2}}+\sqrt{\frac{b^2}{\left(a+b\right)^2}}\)
\(\Leftrightarrow\frac{1}{A}=\left|\frac{a}{a+b}\right|+\left|\frac{b}{a+b}\right|=\frac{a}{a+b}+\frac{b}{a+b}\left(a>2003;b>2003\right)\)
\(\Leftrightarrow\frac{1}{A}=\frac{a+b}{a+b}=1\Leftrightarrow A=1\)
Vậy............................
Bài 2:
\(P=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=\frac{1-\sqrt{5}}{-4}+\frac{\sqrt{5}-\sqrt{9}}{-4}+..+\frac{\sqrt{2001}-\sqrt{2005}}{-4}\)
\(=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(=\frac{1-\sqrt{2005}}{-4}\)
\(=\frac{\sqrt{2005}-1}{4}\)
Dat bieu thuc tren la A
ta co \(\frac{1}{\sqrt{n+2}+\sqrt{n}}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)
ap dung dang thuc tren ta co\(\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}\)
tuong tu ta co \(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\)
.........
\(\frac{1}{\sqrt{2017}+\sqrt{2015}}=\frac{\sqrt{2017}-\sqrt{2015}}{2}\)
ta co
\(A=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+.....+\sqrt{2017}-\sqrt{2015}\right)=\frac{\sqrt{2017}-1}{2}\)
\(Q=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
=> \(Q=\frac{1-\sqrt{5}}{-4}+\frac{\sqrt{5}-\sqrt{9}}{-4}+\frac{\sqrt{9}-\sqrt{13}}{-4}+...+\frac{\sqrt{2001}-\sqrt{2005}}{-4}\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}\right)\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{2005}\right)\)
=> \(Q=\frac{\sqrt{2005}-1}{4}\)
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
ở đây nhé :
www.kichdam.vn
\(A=\frac{1}{\sqrt{2001}+\sqrt{2003}}+\frac{1}{\sqrt{2003}+\sqrt{2005}}+...+\frac{1}{\sqrt{2015}+\sqrt{2017}}\)
Ta có công thức:
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
Áp dụng vào công thức ta có:
\(A=\frac{1}{\sqrt{2001}+\sqrt{2003}}+\frac{1}{\sqrt{2003}+\sqrt{2005}}+...+\frac{1}{\sqrt{2015}+\sqrt{2017}}\)
\(A=\sqrt{2003}-\sqrt{2001}+\sqrt{2005}-\sqrt{2003}+...+\sqrt{2017}-\sqrt{2015}\)
\(A=\sqrt{2017}-\sqrt{2001}\approx0,17848\)