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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 \(3\sqrt{6}< 2\sqrt{15}\)
a) 7 và \(\sqrt{37}+1\)
=7 và 7,08
=>......
b) \(\sqrt{17}-\sqrt{50}-1\)và \(\sqrt{99}\)
=-3,95 và 9,95
=>.....
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
a)A= \(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)=\(\sqrt{6+2\sqrt{3}+2}\)
=> A2=8+2\(\sqrt{3}\)
B=\(\sqrt{3}+1\)=> B2=10+2\(\sqrt{3}\)
=>A>B
\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
1.a)
\(2\sqrt{3}=\sqrt{12}>\sqrt{9}=3.\)
\(3\sqrt{2}=\sqrt{18}>\sqrt{16}=4.\)
Suy ra VT > 7
1.b)
\(\sqrt{16}+\sqrt{25}=4+5=9\)
2.a)
\(\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}\right)^2-6\sqrt{6}+3}=3\sqrt{2}-\sqrt{3}\)
b)\(\sqrt{9-2\sqrt{14}}=\sqrt{\frac{18-4\sqrt{14}}{2}}=\frac{\sqrt{14}-2}{\sqrt{2}}=\sqrt{7}-1\)
Các câu còn lại bạn làm tương tự nhé!
c) \(\sqrt{4-\sqrt{7}}=\frac{1}{\sqrt{2}}.\sqrt{8-2\sqrt{7}}=\frac{1}{\sqrt{2}}\sqrt{7-2\sqrt{7}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{\left(\sqrt{7}-1\right)^2}=\frac{\sqrt{2}\left(\sqrt{7}-1\right)}{2}\)
d) \(\sqrt{4+2\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{4+2\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{4+2\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}-\sqrt{3}+1}=\sqrt{5+\sqrt{3}}\)
b: \(\sqrt{3}-1=\sqrt{4-2\sqrt{3}}\)
mà \(4-3\sqrt{3}< 4-2\sqrt{3}\)
nên \(\sqrt{4-3\sqrt{3}}< \sqrt{3}-1\)
Đề này sai rồi bạn vì \(4-3\sqrt{3}< 0\)