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\(b,\) 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:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Ta có : \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Mà : \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}-\sqrt{2006}}\)
Nến : \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)
\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
\(\left(\sqrt{2005}+\sqrt{2007}\right)^2=4012+2\sqrt{2005.2007}\)
\(=4012+2\sqrt{\left(2016-1\right)\left(2016+1\right)}=4012+2\sqrt{2016^2-1}\)
\(\left(2\sqrt{2006}\right)^2=4012+4012=4012+2\sqrt{2016^2}\)
=>\(\left(\sqrt{2015}+\sqrt{2017}\right)^2< \left(2\sqrt{2016}\right)^2\Rightarrow\sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}\)
Ta có: \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Mà: \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Nên: \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)
=>\(\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)
\(\text{a) Ta có }:\left(\sqrt{7}-\sqrt{2}\right)^2=7-\sqrt{14}+2=9-\sqrt{14}\\ 1^2=1=9-8=9-\sqrt{64}\\ Do\text{ }\sqrt{14}< \sqrt{64}\Rightarrow9-\sqrt{14}>9-\sqrt{64}\\ \Rightarrow\left(\sqrt{7}-\sqrt{2}\right)^2>1^2\\ \Rightarrow\sqrt{7}-\sqrt{2}>1\)
\(\text{b) Ta có: }\left(\sqrt{8}+\sqrt{5}\right)^2=8+\sqrt{160}+5=13+\sqrt{160}\\ \left(\sqrt{7}+\sqrt{6}\right)^2=7+\sqrt{168}+6=13+\sqrt{168}\\ \text{Do }\sqrt{160}< \sqrt{168}\Rightarrow13+\sqrt{160}< 13+\sqrt{168}\\ \Rightarrow\left(\sqrt{8}+\sqrt{5}\right)^2< \left(\sqrt{7}+\sqrt{6}\right)^2\\ \Rightarrow\sqrt{8}+\sqrt{5}< \sqrt{7}+\sqrt{6}\)
\(\text{c) Ta có }:\left(\sqrt{2005}+\sqrt{2007}\right)^2\\ =2005+2\sqrt{2005\cdot2007}+2007\\ =4012+2\sqrt{2005\cdot2007}\\ \left(2\sqrt{2006}\right)^2=4\cdot2006=4012+2\cdot2006\)
\(\text{Lại có }:\sqrt{2005\cdot2007}=\sqrt{\left(2006-1\right)\left(2006+1\right)}=\sqrt{2006^2-1}\\ Do\text{ }\sqrt{2006^2-1}< \sqrt{2006^2}\\ \Rightarrow\sqrt{2005\cdot2007}< 2006\\ \Rightarrow2\sqrt{2005\cdot2007}< 2\cdot2006\\ \Rightarrow4012+2\sqrt{2005\cdot2007}< 4012+2\cdot2006\\ \Rightarrow\left(\sqrt{2005}+\sqrt{2007}\right)^2< \left(2\sqrt{2006}\right)^2\\ \Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
a/ giả sử \(\sqrt{7}-\sqrt{2}< 1\)
\(\Leftrightarrow\sqrt{7}< 1+\sqrt{2}\)
\(\Leftrightarrow 7< 1+2\sqrt{2}+2\)
\(\Leftrightarrow4< 2\sqrt{2}\Leftrightarrow16< 8\left(sai\right)\)
vậy \(\sqrt{7}-\sqrt{2}>1\)
câu b, c bạn làm tương tụ nhé , giả sử một đẳng thức tạm, sau đó bình phương lên rồi làm theo như trên là được nha
Bài này cũng dễ
a, \(\sqrt{7}-\sqrt{2}\) lớn hơn \(1\) . Vì
\(\sqrt{7}-\sqrt{2}=1,231537749\)
\(1=1\)
b, \(\sqrt{8}+\sqrt{5}\) bé hơn \(\sqrt{7}+\sqrt{6}\) . Vì
\(\sqrt{8}+\sqrt{5}=5,064495102\)
\(\sqrt{7}+\sqrt{6}=5,095241054\)
c, \(\sqrt{2005}+\sqrt{2007}\) lớn hơn \(\sqrt{2006}\) . Vì
\(\sqrt{2005}+\sqrt{2007}=89,57677992\)
\(\sqrt{2006}=44,78839135\)