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\(a=\dfrac{8}{\sqrt{2019}+\sqrt{2011}}\)
\(b=\dfrac{8}{\sqrt{19}+\sqrt{11}}\)
Do đó: a<b
a: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)
\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)
mà \(2\sqrt{22}< 15+10\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)
b: \(\left(\sqrt{8}+\sqrt{11}\right)^2=19+2\cdot\sqrt{88}=19+\sqrt{352}\)
\(\left(\sqrt{38}\right)^2=19+19=19+\sqrt{361}\)
mà 352<361
nên \(\sqrt{8}+\sqrt{11}< \sqrt{38}\)
\(a\)
\(\sqrt{11}+\sqrt{19}\)
\(=\)\(\sqrt{11+19}\)
\(=\)\(\sqrt{30}\)
\(=\)\(5,47\)
\(\sqrt{47}\)
\(=6,85\)
\(5,47\)\(< \)\(6,85\)
\(=>\)\(\sqrt{11}+\sqrt{19}\)\(< \)\(\sqrt{47}\)
\(b\)
\(\sqrt{7}+\sqrt{26}+1\)
\(=\)\(\sqrt{7+26}+1\)
\(=\)\(\sqrt{33}+1\)
\(=\)\(5,74+1\)
\(=\)\(6,74\)
\(\sqrt{63}\)
\(=\)\(7,93\)
\(6,74\)\(< \)\(7,93\)
\(=>\)\(\sqrt{7}+\sqrt{26}+1\)\(< \)\(\sqrt{63}\)
Học tốt!!!
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 \(3\sqrt{6}< 2\sqrt{15}\)