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A= ( \(\sqrt{1}\)+\(\sqrt{2}\)+\(\sqrt{3}\) ) + (\(\sqrt{20}\) + \(\sqrt{40}\) + \(\sqrt{60}\))
= (1+1,4+1,7)+(4,4+6,3+7,7)
= 4,1+18,4
=22,5
Ta sẽ chứng minh 1 bđt sau:
\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
\(\Rightarrow a+2\sqrt{ab}+b\ge a+b\)
\(\Rightarrow a+2\sqrt{ab}+b-a-b\ge0\)
\(\Rightarrow2\sqrt{ab}\ge0\) *đúng*
Dấu "=" xảy ra khi: \(ab=0\)
Trở lại bài toán,vì không có thừa số nào bằng 0,nên ta dễ dàng có: \(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
Hay \(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}=\left(\sqrt{1}+\sqrt{20}\right)+\left(\sqrt{40}+\sqrt{2}\right)+\left(\sqrt{60}+\sqrt{3}\right)>\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}=A\)
\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{10}{3^2}.\frac{20}{4^2}.......\frac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\Rightarrow\frac{1.2}{1.1}.\frac{2.3}{2.2}.\frac{3.4}{3.3}.\frac{4.5}{4.4}......\frac{10.11}{10.10}\left(x-2\right)=-20x-20+60\)
\(\Rightarrow\frac{1.2.3.4.....10}{1.2.3.4.....10}.\frac{2.3.4.5.....11}{1.2.3.4.....10}\left(x-2\right)=-20x+40\)
\(\Rightarrow11\left(x-2\right)=-20x+40\)
\(\Rightarrow11x-22=-20x+40\)
\(\Rightarrow11x+20x=22+40\)
\(\Rightarrow31x=62\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{20}.20.21:2=\frac{2}{2}+\frac{3}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+...+21}{2}=\frac{230}{2}=115\)
B = 1+1/2×(2×3/2)+1/3×(3×4/2)+1/4×(4×5/2)+...+1/20×(20×21/2)=1+3/2+4/2+...+21/2=1/2×(2+3+4+...+21=1/2×(2+3+4+...+21)=1/2×(21×22/2-1)=115
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)\(=1+\dfrac{1}{2}.2.3:2+\dfrac{1}{3}.3.4:2+...+\dfrac{1}{20}.20.21:2\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{21}{2}\)
\(=\dfrac{2+3+...+21}{2}\)
\(=\dfrac{230}{2}\)
\(=115\)