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\(\text{a) Ta có }:\left(\sqrt{7}-\sqrt{2}\right)^2=7-\sqrt{14}+2=9-\sqrt{14}\\ 1^2=1=9-8=9-\sqrt{64}\\ Do\text{ }\sqrt{14}< \sqrt{64}\Rightarrow9-\sqrt{14}>9-\sqrt{64}\\ \Rightarrow\left(\sqrt{7}-\sqrt{2}\right)^2>1^2\\ \Rightarrow\sqrt{7}-\sqrt{2}>1\)
\(\text{b) Ta có: }\left(\sqrt{8}+\sqrt{5}\right)^2=8+\sqrt{160}+5=13+\sqrt{160}\\ \left(\sqrt{7}+\sqrt{6}\right)^2=7+\sqrt{168}+6=13+\sqrt{168}\\ \text{Do }\sqrt{160}< \sqrt{168}\Rightarrow13+\sqrt{160}< 13+\sqrt{168}\\ \Rightarrow\left(\sqrt{8}+\sqrt{5}\right)^2< \left(\sqrt{7}+\sqrt{6}\right)^2\\ \Rightarrow\sqrt{8}+\sqrt{5}< \sqrt{7}+\sqrt{6}\)
\(\text{c) Ta có }:\left(\sqrt{2005}+\sqrt{2007}\right)^2\\ =2005+2\sqrt{2005\cdot2007}+2007\\ =4012+2\sqrt{2005\cdot2007}\\ \left(2\sqrt{2006}\right)^2=4\cdot2006=4012+2\cdot2006\)
\(\text{Lại có }:\sqrt{2005\cdot2007}=\sqrt{\left(2006-1\right)\left(2006+1\right)}=\sqrt{2006^2-1}\\ Do\text{ }\sqrt{2006^2-1}< \sqrt{2006^2}\\ \Rightarrow\sqrt{2005\cdot2007}< 2006\\ \Rightarrow2\sqrt{2005\cdot2007}< 2\cdot2006\\ \Rightarrow4012+2\sqrt{2005\cdot2007}< 4012+2\cdot2006\\ \Rightarrow\left(\sqrt{2005}+\sqrt{2007}\right)^2< \left(2\sqrt{2006}\right)^2\\ \Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
\(b,\) 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: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
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 : \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)
\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
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
\(\sqrt{2005}+\sqrt{2007}=89,57677992\)
\(\sqrt{2006}=44,78839135\)