Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1-2x\ge0\Leftrightarrow2x\le1\Leftrightarrow x\le\frac{1}{2}\)
\(x-\sqrt{1-2x}>0\)
X>0
P2\(=\left(\frac{1-A\sqrt{A}}{1-\sqrt{A}}+\sqrt{A}\right).\left(\frac{1-\sqrt{A}}{1-A}\right)^2\)\(=\left(\frac{1-A\sqrt{A}+\sqrt{A}-A}{1-\sqrt{A}}\right).\frac{\left(1-\sqrt{A}\right)^2}{\left(1-A\right)^2}\)\(=\frac{\left(\sqrt{A}+1\right)\left(1-A\right)}{1-\sqrt{A}}.\frac{\left(1-\sqrt{A}\right)^2}{\left(1-\sqrt{A}\right)^2\left(1+\sqrt{A}\right)^2}\)
\(=\left(\sqrt{A}+1\right)^2.\frac{1}{\left(1+\sqrt{A}\right)^2}=1\)
\(A=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{2}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+3\sqrt{x}+2+2\sqrt{2}.\sqrt{x}-4\sqrt{2}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}\left(\sqrt{2}-1\right)-4\sqrt{2}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\sqrt{2}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+2\sqrt{2}}{\sqrt{x}+2}\)
a, \(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(\Rightarrow\) \(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
\(\Rightarrow\) \(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(\Rightarrow\) \(2S=1-\frac{1}{2017}\)
\(\Rightarrow\) \(2S=\frac{2016}{2017}\)
\(\Rightarrow\) \(S=\frac{1008}{2017}\)
Ta có công thức tổng quát: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)(*)
Áp dụng (*), ta được: \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+...+\left(\sqrt{100}-\sqrt{99}\right)=\sqrt{100}-\sqrt{1}=9\left(đpcm\right)\)
Trục căn thức ở mẫu :
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}+...+\frac{\sqrt{99}-\sqrt{100}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{99}-\sqrt{100}\right)}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-\sqrt{1}\)
\(=10-1=9\)
=> đpcm
Dạng tổng quát ta càn chứng minh \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\)
Ta có \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\frac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\left(\frac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}\)
\(=\frac{a^2+ab+b^2}{ab\left(a+b\right)}=\frac{1}{b}+\frac{b}{a\left(a+b\right)}=\frac{1}{b}+\frac{1}{a}-\frac{1}{a+b}\left(đpcm\right)\)
Áp dụng dạng trên ta được
\(D=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{99}-\frac{1}{100}\)
\(D=100-\frac{1}{100}=\frac{9999}{100}\)
Xét biểu thức \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)với a > 0
\(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}=\left[\frac{a^2+a+1}{a\left(a+1\right)}\right]^2\)Do a > 0 nên A > 0 và \(A=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Do đó \(D=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)=99+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100-\frac{1}{100}=99,99\)
Đặt \(\hept{\begin{cases}\sqrt{1+\frac{\sqrt{3}}{2}}=a\\\sqrt{1-\frac{\sqrt{3}}{2}}=b\end{cases}}\)
\(\Rightarrow a^2+b^2=2;ab=\frac{1}{2};a-b=1\)
\(\Rightarrow\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{a^2}{1+a}+\frac{b^2}{1-b}\)
\(=\frac{a^2+b^2-ab\left(a-b\right)}{1-ab+\left(a-b\right)}=\frac{2-\frac{1}{2}.1}{1-\frac{1}{2}+1}=1\)
\(=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{\left(\sqrt{0.75}+\sqrt{0.25}\right)^2}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{\left(\sqrt{0.75}-\sqrt{0.25}\right)^2}}\)
\(=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{0.75}+\sqrt{0.25}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{0.75}+\sqrt{0.25}}\)
TRỤC CĂN THỨC Ở MẪU TA ĐƯỢC
\(=\frac{9+4\sqrt{3}}{33}+\frac{3-\sqrt{3}}{6}\)
Quy đồng ta được
\(=\frac{17-\sqrt{3}}{22}\)
TICK CHO MÌNH NHA BẠN