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Trước hết ta chứng minh bất đẳng thức tổng quát : với n là là số tự nhiên lớn hơn 1 thì :
\(2\sqrt{n-2< 1+1\sqrt{2}+1\sqrt{3}+....+1\sqrt{n}< 2\sqrt{n}-12n-2< 1+12+13+...+1n< 2n-1\left(\cdot\right)\left(\cdot\right)}\)Xét số hạng thứ kk trong dãy : (2 bé hơn hoặc k bé hơn hoặc bằng n ).(2 bé hơn hoặc bằng k bé hơn hoặc bằng n )
Ta có : \(1\sqrt{k>2\sqrt{k}+\sqrt{k}+1=2\left(\sqrt{k}+1-\sqrt{k}\right)1k>2k+k+1=2\left(k+1-k\right)v\text{à}}1\sqrt{k}< 2\sqrt{k}+\sqrt{k}-1=2\left(\sqrt{k}-\sqrt{k}-1\right)1k< 2k+k-1\)\(=2\left(k-k-1\right)\)
Do đó : \(1+1\sqrt{2}+...+1\sqrt{n}>2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+....+\sqrt{n}+1-\sqrt{n}\right)=2\left(\sqrt{n}+1-1\right)>2\sqrt{n}-21+12+.....+1n\)\(>2\left(2-1+3-2+...+n+1-n\right)=2\left(n+1-1\right)>2n-2v\text{à}1+1\sqrt{2}+.....+1\sqrt{n}< 1+2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{n}-\sqrt{n}-1\right)\)\(=1+2\left(\sqrt{n}-1\right)=2\sqrt{n}-11+12+...+1n< 1+2\left(2-1+3-2+...+n-n-1\right)=1+2\left(n-1\right)=2n-1\)Đến đây áp dụng (*)(*) với n=100n=100 thì 19<a<2019<a<20 nên a không phải là số tự nhiên
a) Ta có: \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=\sqrt{2}\left(3+4\cdot2-3\right)\)
\(=8\sqrt{2}\)
b) Ta có: \(\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\)
\(=\sqrt{3}\left(1-\frac{1}{3}\cdot\sqrt{9}+2\cdot\sqrt{169}\right)\)
\(=\sqrt{3}\left(1-1+26\right)\)
\(=26\sqrt{3}\)
c) Ta có: \(\sqrt{25a}+\sqrt{49a}-\sqrt{64a}\)
\(=\sqrt{25}\cdot\sqrt{a}+\sqrt{49}\cdot\sqrt{a}-\sqrt{64}\cdot\sqrt{a}\)
\(=\sqrt{a}\left(5+7-8\right)\)
\(=4\sqrt{a}\)
d) Ta có: \(-\sqrt{36b}-\frac{1}{3}\sqrt{54b}+\frac{1}{5}\sqrt{150b}\)
\(=-\sqrt{6b}\cdot\sqrt{6}-\frac{1}{3}\cdot\sqrt{6b}\cdot\sqrt{9}+\frac{1}{5}\cdot\sqrt{6b}\cdot\sqrt{25}\)
\(=-\sqrt{6b}\left(\sqrt{6}+1-1\right)\)
\(=-\sqrt{6b}\cdot\sqrt{6}=-6\sqrt{b}\)
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
a, \(=7\sqrt{2}-6\sqrt{2}+\frac{1}{2}.2\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b, \(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}=4\sqrt{a}-5\sqrt{10a}\)
c, \(=6+\sqrt{15}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
Rút gọn
a) Ta có: \(\sqrt{98}-\sqrt{72}+\frac{1}{2}\sqrt{8}\)
\(=\sqrt{2}\left(\sqrt{49}-\sqrt{36}+\frac{1}{2}\sqrt{4}\right)\)
\(=\sqrt{2}\left(7-6+\frac{1}{2}\cdot2\right)\)
\(=\sqrt{2}\left(1+1\right)=2\sqrt{2}\)
b) Ta có: \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)
\(=\sqrt{a}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)
\(=\sqrt{a}\left(4+4\sqrt{10}-9\sqrt{10}\right)\)
\(=\sqrt{a}\left(4-5\sqrt{10}\right)\)
\(=4\sqrt{a}-5\sqrt{10a}\)
c) Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)
\(=6+\sqrt{15}-\sqrt{60}\)
\(=6-\sqrt{15}\)
a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)
b)\(\sqrt{125a^2}=\sqrt{5^2.5.a^2}=5.\left|a\right|\sqrt{5}=-5a\sqrt{5}\) ( vì a<0)
c)\(-\sqrt{500.162}=-\sqrt{10^2.5.9^2.2}=-10.9\sqrt{5.2}=-90\sqrt{10}\)
d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{1}{3}.15.\left|a\right|=\frac{15a}{3}\) ( a>0)
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)