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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)
\(\Leftrightarrow x=216-1=215\)
1/2.(1/3+1/6+1/10+...+1/x(x+1))=1/2.2016/2018
1/6+1/12+1/20+...+1/x(x+1)=504/1009
1/2.3+1/3.4+1/4.5+...+1/x(x+1)=504/1009
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=504/1009
1/2-1/x+1=504/1009
x-1/2(x+1)=504/1009
-> 1009(x-1)=504.2(x+1)
1009x-1009=1008x+1008
1009x-1008x=1008+1009
->x=2017
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2016}{2018}\)
\(A=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{x\left(x+1\right):2}\)
\(A=\frac{1}{2\left(2+1\right)}\cdot2+\frac{1}{3\left(3+1\right)}\cdot2+...+\frac{1}{x\left(x+1\right)}.2=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=1-\frac{1}{x+1}=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2016}{2018}=\frac{1}{1009}\)
\(\Rightarrow x+1=1009\Rightarrow x=1008\)
\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{9.9}\)
\(N\)bé hơn \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}=N_1\)
\(N_1=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\) \((1)\)
\(N\)lớn hơn \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}=N_2\)
\(\Rightarrow N_2=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{5}{10}-\frac{1}{10}=\frac{2}{5}\) \((2)\)
Từ \((1)\)và \((2)\)suy ra ; \(\frac{2}{5}\)bé hơn N bé hơn \(\frac{8}{9}\)
Học tốt
Nhớ kết bạn với mình
Theo bđt cô si ta có : \(x+y\ge2\sqrt{xy}\) <=> \(1\ge2\sqrt{xy}\)
=> \(\sqrt{xy}\le\frac{1}{2}\) <=> \(\sqrt{\frac{1}{xy}}\ge2\)
Theo bđt cô si : \(P=\frac{a^2}{x}+\frac{b^2}{y}\ge2\sqrt{\frac{a^2b^2}{xy}}=2ab\sqrt{\frac{1}{xy}}=2ab.2=4ab\)
Vậy giá trị nhỏ nhất của P=4ab khi x=y=1/2
a) \(\left|x+3\right|:\left(-15\right)=\frac{1}{3}\)
\(\left|x+3\right|=-5\)
=> không tìm được x
b) \(\left|4,5-2x\right|.\left(-1\frac{4}{7}\right)=-\frac{11}{14}\)
\(\left|4,5-2x\right|=2\)
TH1: 4,5 - 2x = 2
2x = 2,5
x = 1,25
TH2: 4,5 - 2x = -2
2x = 6,5
x = 3,25
KL:...
Trả lời
x:(1/2+1/3+1/6)=2019
x:6/6 =2019
x:1 =2019
=>x =2019.1
=>x =2019
Học tốt !
x : (\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{6}\)) = 2019
x : 1 = 2019
x = 2019