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Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\)
\(A=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)\)
\(>1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
\(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(=1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
\(>1+2\times\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=1+2\times1\)
\(=1+2=3=B\)
\(\Rightarrow A>B\)
Học tốt
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+.........+\frac{1}{3^{100}}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+.........+\frac{1}{3^{100}}-\left(\frac{1}{3}+\frac{1}{3^2}+.......+\frac{1}{3^{99}}\right)=1+\frac{1}{3}\)
\(\Rightarrow2A=1+\frac{1}{3}\Rightarrow A=\left(1+\frac{1}{3}\right):2\)
=>3A=1/3^2+1/3^3+1/3^4+...+1/3^100
=>3A-A=(1/3^2+1/3^3+1/3^4+...+1/3^100) - (1/3+1/3^2+1/3^3+...+1/3^99)
=>2A=1/3^100-1/3
=>A=(\(\frac{1}{3^{100}}\)- \(\frac{1}{3}\)):2
Li ke mình nha!
\(3A=\frac{1}{1}+\frac{2}{3}+\frac{3}{3^2}+....+\frac{101}{3^{100}}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}< 1+\frac{1}{3}+...+\frac{1}{3^{100}}< \frac{3}{2}\Rightarrow A< \frac{3}{4}\)
chủ yếu là cách làm thôi, có gì bạn tự tính
1
a) Ta có\(\frac{31}{40}=\frac{31.6}{40.6}=\frac{186}{240}\)
Vì \(240< 241\)
nên\(\frac{286}{240}>\frac{286}{241}\)
Vậy\(\frac{31}{40}>\frac{286}{240}\)
b)Ta có\(\frac{411}{911}=\frac{911-500}{911}=1-\frac{500}{911}\)
\(\frac{41}{91}=\frac{91-50}{91}=1-\frac{50}{91}=1-\frac{500}{910}\)
Vì \(\frac{500}{911}< \frac{500}{910}\)nên\(1-\frac{500}{911}>1-\frac{500}{910}\)
Vậy \(\frac{411}{911}>\frac{41}{91}\)
Trong day 1/√1+...+1/√100 thi 1/√100 la be nhat
Ma 1/√100=1/10
A>1/10+1/10+...+1/10
Ma day tren co 100 so
=> A>100/10=10
=> A>10