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a)1/7\(\sqrt{51}\)=\(\sqrt{\frac{51}{49}}\);1/9\(\sqrt{150}=\sqrt{\frac{150}{81}}=\sqrt{\frac{50}{27}}\)
\(\frac{51}{49}=1+\frac{1}{49}+\frac{1}{49}\);\(\frac{50}{27}=1+\frac{23}{27}>1+\frac{23}{36}>\)\(1+\frac{2}{36}=1+\frac{1}{36}+\frac{1}{36}\)
1/49<1/36 nên 51/49<50/27 =>1/7\(\sqrt{51}\)<1/9\(\sqrt{150}\)
b) \(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}\)+\(\sqrt{2015}\)
=>\(\frac{1}{\sqrt{2017}+\sqrt{2016}}< \)\(\frac{1}{\sqrt{2016}+\sqrt{ }2015}\) <=> \(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}\)-\(\sqrt{2015}\)
Lời giải:
a)
Ta có: \(\frac{1}{7}\sqrt{51}< \frac{1}{7}\sqrt{64}=\frac{8}{7}\)
\(\frac{1}{9}\sqrt{150}> \frac{1}{9}\sqrt{144}=\frac{12}{9}=\frac{4}{3}=\frac{8}{6}> \frac{8}{7}\)
Do đó: \(\frac{1}{7}\sqrt{51}< \frac{1}{9}\sqrt{150}\)
b)
\(\sqrt{2017}-\sqrt{2016}=\frac{2017-2016}{\sqrt{2017}+\sqrt{2016}}=\frac{1}{\sqrt{2017}+\sqrt{2016}}< \frac{1}{\sqrt{2016}+\sqrt{2015}}\)
\(\sqrt{2016}-\sqrt{2015}=\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\)
Do đó:
\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)
<=> \(\dfrac{\sqrt{51}}{7}với\dfrac{\sqrt{150}}{9}\)
<=> \(9\sqrt{51}với7\sqrt{150}\)
<=> \(\sqrt{4131}với\sqrt{7350}\)
=> \(\sqrt{4131}< \sqrt{7350}\)
=> \(\dfrac{1}{7}\sqrt{51}< \dfrac{1}{9}\sqrt{150}\)
Ta có:
\(\sqrt{2016}-\sqrt{2017}=\frac{\left(\sqrt{2016}-\sqrt{2017}\right)\left(\sqrt{2016}+\sqrt{2017}\right)}{\sqrt{2016}+\sqrt{2017}}\)
\(=\frac{2016-2017}{\sqrt{2016}+\sqrt{2017}}=-\frac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2017}-\sqrt{2018}=\frac{\left(\sqrt{2017}-\sqrt{2018}\right)\left(\sqrt{2017}+\sqrt{2018}\right)}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{2017-2018}{\sqrt{2017}+\sqrt{2018}}=-\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
Ta thấy rằng:
\(\sqrt{2018}>\sqrt{2016}\)
\(\Leftrightarrow\sqrt{2017}+\sqrt{2018}>\sqrt{2016}+\sqrt{2017}\)
\(\Leftrightarrow\frac{1}{\sqrt{2017}+\sqrt{2018}}< \frac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\Leftrightarrow-\frac{1}{\sqrt{2017}+\sqrt{2018}}>-\frac{1}{\sqrt{2016}+\sqrt{2017}}\)
Vậy \(\sqrt{2017}-\sqrt{2018}>\sqrt{2016}-\sqrt{2017}\)
~ ~ ~
\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\)
\(=\sqrt{\dfrac{37}{4}-\sqrt{\left(3\sqrt{5}+2\right)^2}}\)
\(=\sqrt{\dfrac{29}{4}-3\sqrt{5}}\)
\(=\sqrt{\dfrac{29-12\sqrt{5}}{4}}\)
\(=\sqrt{\dfrac{\left(2\sqrt{5}-3\right)^2}{4}}\)
\(=\dfrac{\sqrt{5}}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}\left(\sqrt{5}-\dfrac{3}{2}\right)\)
\(>\sqrt{5}-\dfrac{3}{2}=B\)
~ ~ ~
\(C=\dfrac{16\sqrt{36}-20\sqrt{48}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-80\sqrt{3}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-70\sqrt{3}}{2\sqrt{3}}\)
\(=16\sqrt{3}-35\)
\(>16\sqrt{3}-36=B\)
~ ~ ~
b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)
nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)