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\(\frac{2016}{\sqrt{2016}}=\sqrt{2016}\)
\(\frac{2017}{\sqrt{2017}}=\sqrt{2017}\)
=> Bằng nhau
\(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}=\left(\frac{2016}{\sqrt{2017}}-\sqrt{2017}\right)+\left(\frac{2017}{\sqrt{2016}}-\sqrt{2016}\right)\)
\(=\frac{2016-2017}{\sqrt{2017}}+\frac{2017-2016}{\sqrt{2016}}=\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
vì \(2016< 2017\Rightarrow\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)
\(\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}>0\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2016}+\sqrt{2017}\)
\(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\sqrt[3]{\left(1-\sqrt{3}\right)\left(\sqrt{3}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{3}\right)^3}\)=1-\(\sqrt{3}\)
\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}=\sqrt[3]{\left(1-\sqrt{5}\right)\left(\sqrt{5}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=1-\(\sqrt{5}\)
Ta thấy \(\sqrt{5}>\sqrt{3}\)nên 1-\(\sqrt{3}\)>\(1-\sqrt{5}\)
Vậy \(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}\)>\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
Ta có
\(\left(2+\sqrt{3}\right)^2=2^2+2\cdot2\cdot\sqrt{3}+3=7+4\sqrt{3}\)
\(\Rightarrow2+\sqrt{3}=\sqrt{7+4\sqrt{3}}\)
Ta có \(7+4\sqrt{3}>5+4\sqrt{3}\)
\(\Leftrightarrow\sqrt{7+4\sqrt{3}}>\sqrt{5+4\sqrt{3}}\)
\(\Rightarrow2+\sqrt{3}>\sqrt{5+4\sqrt{3}}\)
Bài 2:
a: ĐKXĐ: 3x-7>=0
hay x>=7/3
b: ĐKXĐ: \(2-5x\ge0\)
hay x<=2/5
c: ĐKXĐ: \(\dfrac{-3}{x-5}\ge0\)
=>x-5<0
hay x<5
d: ĐKXĐ: \(5x^2-x-4\ge0\)
\(\Leftrightarrow5x^2-5x+4x-4\ge0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+4\right)\ge0\)
=>x>=1 hoặc x<=-4/5
e: ĐKXĐ: \(9-x^2\ge0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\le0\)
=>-3<=x<=3
f: ĐKXĐ: \(x^2-1\ge0\)
=>(x-1)(x+1)>=0
=>x>=1 hoặc x<=-1
Lời giải:
a)
\(\frac{4}{\sqrt{10}}(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})=\frac{4}{\sqrt{20}}(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}})\)
\(=\frac{4}{2\sqrt{5}}(\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}})=\frac{2}{\sqrt{5}}[\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}]\)
\(=\frac{2}{\sqrt{5}}(\sqrt{5}+1+\sqrt{5}-1)=\frac{2}{\sqrt{5}}.2\sqrt{5}=4\)
b)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
c)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+8\sqrt{3}+18}=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(3+1+2\sqrt{3})+2}\)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(\sqrt{3}+1)^2+2}\)
\(=\sqrt{(2\sqrt{3}+2)^2+(\sqrt{2})^2+2.(2\sqrt{3}+2).\sqrt{2}}\)
\(=\sqrt{(2\sqrt{3}+2+\sqrt{2})^2}=2\sqrt{3}+2+\sqrt{2}\)
1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)
\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)
\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)
2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)
\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)
Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)
3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)
Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)