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\(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}\)\(=\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+...+\sqrt{10.11}\)
\(< \frac{1+2}{2}+\frac{2+3}{2}+\frac{3+4}{2}+...+\frac{10+11}{2}\)\(=\frac{1}{2}\left[\left(1+2+3+...+10\right)+\left(2+3+4+...+11\right)\right]\)\(=\frac{1}{2}\left(\frac{11.10}{2}+\frac{13.10}{2}\right)=\frac{1}{2}\left(55+65\right)=60\)
Vậy \(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}< 60.\)
\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)
Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)
Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)
\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)
Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)
Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
Do \(\sqrt{1}=1;\sqrt{2}+\sqrt{3}+\sqrt{4}< 3.\sqrt{4}=6\)\(;\sqrt{5}+\sqrt{6}+...+\sqrt{9}< 5.\sqrt{9}=15\)
\(\Rightarrow\sqrt{1}+\sqrt{2}+...+\sqrt{9}< 1+6+15=22\)(1)
Cung co:\(5.\sqrt{5}>5.\sqrt{4}=10\)\(\Rightarrow5.\sqrt{5}+12>10+12=22\)(2)
Tu (1) va (2) =>....
a: \(\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\)
\(\left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)
mà \(-2\sqrt{105}>-2\sqrt{120}\)
nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b: \(\left(\sqrt{2}+\sqrt{8}\right)^2=10+2\cdot4=16=12+4\)
\(\left(3+\sqrt{3}\right)^2=12+6\sqrt{3}\)
mà \(4< 6\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{8}< 3+\sqrt{3}\)
\(A=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)
\(A< \sqrt{2,25}+\sqrt{6,25}+\sqrt{12,25}+\sqrt{20,25}+\sqrt{30,25}+\sqrt{42,25}=24=B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
