Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Mình chỉ viết CT tổng quát thôi nha rồi bạn tự thay vào
a, \(\frac{1}{\sqrt{n}(n+1)+n\sqrt{n+1} }=\frac{1}{\sqrt{n(n+1)( }\sqrt{n}+\sqrt{n+1}} =\frac{\sqrt{n+1}-\sqrt{n} }{\sqrt{n}\sqrt{n+1} } =\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } \)
b,\(\frac{1}{\sqrt{n}+\sqrt{n+1} }=\frac{\sqrt{n+1}-\sqrt{n} }{1}= \sqrt{n+1}-\sqrt{n} \)
Bài này mới học xong nè =)))
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+.....+\dfrac{1}{\sqrt{2005}+\sqrt{2006}}\)
\(=\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{3}-\sqrt{4}}{-1}+....+\dfrac{\sqrt{2005}-\sqrt{2006}}{-1}\)
\(=\dfrac{\sqrt{2}-\sqrt{2006}}{-1}=\sqrt{2006}-\sqrt{2}\)
Sửa đề:
\(VP=\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)
Ta có: \(2005^2+1=\left(2005+1\right)^2-2.2005.1=2006^2-2.2005\)
\(\Rightarrow VP=\sqrt{2006^2-2.2005+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)
\(=\sqrt{\left(2006-\dfrac{2005}{2006}\right)^2}+\dfrac{2005}{2006}\)
\(=2006-\dfrac{2005}{2006}+\dfrac{2005}{2006}=2006\)
Phương trình đã cho tương đương
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=2006\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2006\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)
Đến đây thì tự xét trường hợp và giải tìm nghiệm, bài này không cần điều kiện nhé
\(\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}=\dfrac{1}{2006}\sqrt{2006^2+2005^2+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2006-2005\right)^2+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{1+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2005.2006+1\right)^2}=\dfrac{2005.2006+1}{2006}=2005+\dfrac{1}{2006}\)
Phương trình tương đương:
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2005+\dfrac{1}{2006}+\dfrac{2005}{2006}\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)
TH1: \(x\ge2\): \(x-1+x-2=2006\Rightarrow2x=2009\Rightarrow x=\dfrac{2009}{2}\)
TH2: \(x\le1\) : \(1-x+2-x=2006\Rightarrow-2x=2003\Rightarrow x=\dfrac{-2003}{2}\)
TH3: \(1< x< 2:\) \(x-1+2-x=2006\Rightarrow3=2006\) (vô nghiệm)
Vậy \(\left[{}\begin{matrix}x=\dfrac{2009}{2}\\x=\dfrac{-2003}{2}\end{matrix}\right.\)
Áp dụng bđt AM-GM cho 2 số không âm ta có:
\(\dfrac{1}{\sqrt{1.2006}}>\dfrac{1}{\dfrac{1+2006}{2}}=\dfrac{2}{2007}\)
TT: \(\dfrac{1}{\sqrt{2.2005}}>\dfrac{2}{2007}\)
...
\(\dfrac{1}{\sqrt{2006.1}}>\dfrac{2}{2007}\)
Cộng vế với vế ta được:
\(S>\dfrac{2}{2007}.2006\)
\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)
\(A=\sqrt{9}-\sqrt{1}=3-1=2\)
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
\(P=\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}\)
\(=\dfrac{\sqrt{1}+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}\)
\(=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)
\(=-1+\sqrt{9}=-1+3=2\)
Học tốt !!
\(\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\sqrt{k}+\sqrt{k+1}}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =\dfrac{\sqrt{k}+\sqrt{k+1}}{k-k-1}=-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow P=\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ =-\sqrt{1}+\sqrt{9}=-1+3=4\)
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
Tiếp phần b ( do máy lag) :3
Cộng 2 vế với nhau, ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\\ =1-\dfrac{1}{\sqrt{2007}}\)