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Ta có \(\sqrt{8}+3< \sqrt{9}+3=3+3=6\)
=> \(\sqrt{8}+3< 6\)
Ta có \(\sqrt{48}< \sqrt{49};\sqrt{35}< \sqrt{36}\)
=> \(\sqrt{48}+\sqrt{35}< \sqrt{49}+\sqrt{46}\)
=> \(\sqrt{48}+\sqrt{35}< 13\)
=> \(\sqrt{48}< 13-\sqrt{35}\)
c) Ta có \(-\sqrt{19}< -\sqrt{17}\)
=> \(\sqrt{31}-\sqrt{19}< \sqrt{31}-\sqrt{17}\)
=> \(\sqrt{31}-\sqrt{19}< \sqrt{36}-17=6-\sqrt{17}\)
d) Ta có \(9=\sqrt{81}\Leftrightarrow\sqrt{81}>\sqrt{80}\);
\(-\sqrt{58}>-\sqrt{59}\)
=> \(\sqrt{81}-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
<=> \(9-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
a) Ta có: \(\frac{1}{5}\sqrt{150}=\frac{1}{5}\cdot5\sqrt{6}=\sqrt{6}=\frac{1}{3}\cdot\sqrt{6\cdot9}=\frac{1}{3}\sqrt{54}>\frac{1}{3}\sqrt{51}\)
b) Ta có: \(\frac{1}{2}\sqrt{6}=\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}=6\sqrt{\frac{1}{2}}\)
a) Vì \(5,\left(6\right)< 6\)\(\Rightarrow\)\(\frac{51}{9}< \frac{150}{25}\)
\(\Rightarrow\)\(\sqrt{\frac{51}{9}}< \sqrt{\frac{150}{25}}\)
\(\Rightarrow\)\(\frac{1}{3}\sqrt{51}< \frac{1}{5}\sqrt{150}\)
b) Vì \(1,5< 18\)\(\Rightarrow\)\(\frac{6}{4}< \frac{36}{2}\)
\(\Rightarrow\)\(\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}\)
\(\Rightarrow\)\(\frac{1}{2}\sqrt{6}< 6\sqrt{\frac{1}{2}}\)
a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)
\(\sqrt{6}< \sqrt{6,25}=2,5\);
\(\sqrt{12}< \sqrt{12,25}=3,5\);
\(\sqrt{20}< \sqrt{20,25}=4,5\)
=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)
Vậy P < 12
Answer:
ý a, tham khảo bài làm của @xyzquynhdi
\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{5}+2\sqrt{3}\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
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) \(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}\)+\(\sqrt{2015}\)
=>\(\frac{1}{\sqrt{2017}+\sqrt{2016}}< \)\(\frac{1}{\sqrt{2016}+\sqrt{ }2015}\) <=> \(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}\)-\(\sqrt{2015}\)
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 \(\sqrt{2017}-\sqrt{2018}>\sqrt{2016}-\sqrt{2017}\)
Đặt \(A=\left(\sqrt{2018}+\sqrt{2020}\right)\)
\(\Rightarrow A^2=2018+2\sqrt{2018.2020}+2020=4038+\sqrt{4.2018.2020}=4038+\sqrt{4.\left(2019^2-1\right)}\)
Đặt \(B=2\sqrt{2019}=\sqrt{4.2019}\)
\(B^2=4.2019=2.2019+2.2019=4038+\sqrt{4.2019^2}\)
=> \(\sqrt{4.2019^2}>\sqrt{4.\left(2019^2-1\right)}\)
\(\Rightarrow A>B\Leftrightarrow\sqrt{2018}+\sqrt{2020}>2\sqrt{2019}\)
Ta có:
\(\left(\sqrt{3+\sqrt{20}}\right)^2-\left(\sqrt{5+\sqrt{5}}\right)^2\)
\(=3+\sqrt{20}-5-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
Ta thấy: \(5>4\Rightarrow\sqrt{5}>\sqrt{4}\Rightarrow\sqrt{5}>2\)
Do đó : hiệu trên >0
Suy ra : \(\sqrt{3+\sqrt{20}}>\sqrt{5+\sqrt{5}}\)
\(\sqrt{27}-\sqrt{35}>\sqrt{25}-\sqrt{36}=5-6=-1\)
\(6-\sqrt{51}< 6-\sqrt{49}=6-7=-1\)
=> \(\sqrt{27}-\sqrt{35}>6-\sqrt{51}\)