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Ta có :
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
Chúc bạn học tốt ~
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(\Leftrightarrow A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{49}{100}\)
Vậy A=\(\frac{49}{100}\)
#)Giải :
\(\frac{-5}{12}< \frac{a}{5}< \frac{1}{4}\Leftrightarrow\frac{-25}{60}< \frac{12a}{60}< \frac{15}{60}\Leftrightarrow-25< 12a< 15\)
\(\Leftrightarrow12a\in\left\{\pm12;-24\right\}\)
\(\Leftrightarrow a\in\left\{\pm1;2\right\}\)
Bài giải
Ta có :
\(-\frac{5}{12}< \frac{a}{5}< \frac{1}{4}\)
\(\Leftrightarrow\text{ }-\frac{25}{60}< \frac{12a}{60}< \frac{15}{60}\) \(\Rightarrow\text{ }-25< 12a< 15\)
\(\Rightarrow\text{ }-1,25< a< 1,25\)
\(\text{Do }a\in Z\text{ }\Rightarrow\text{ }x\in\left\{-1\text{ ; }0\text{ ; }1\right\}\)
Ta có: A=1/11+1/12+1/13+...+1/30
=(1/11+1/12+1/13+..+1/20)+(1/21+1/22+1/23+...+1/30)
\(\Rightarrow\)A<(1/10+1/10+1/10+...+1/10)+(1/20+1/20+1/20+...1/20)
\(\Rightarrow\)A<(1/10)*10+(1/20)*10
\(\Rightarrow\)A<1+1/2
\(\Rightarrow\)A<3/2<11/6
\(Ta\)\(có\)\(\frac{x}{5}\)\(=\frac{-12}{20}\)
\(\Rightarrow\frac{-12}{20}=\frac{-12:4}{20:4}\)
\(\Rightarrow\frac{-12}{20}=\frac{-3}{5}\)
\(\Rightarrow x=-3\)
Vậy x=-3
a) \(x=\left(-12\right).5:20=-3\)
Vậy x = -3
b) \(y=\left(-66\right).2:11=-12\)
Vậy x = -12
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A< 1-\frac{1}{10}=\frac{9}{10}\)
\(=>A>\frac{65}{132}\)
Ta có
\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}};\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}\)
\(\Rightarrow\frac{1}{3^4}< \frac{1}{4^3}\left(3^4>4^3\right)\\
\Rightarrow\frac{1}{3^{400}}< \frac{1}{4^{300}}\)
A=24x27-23/24+23x27
A=(23+1)x27-23/24+23x27
A=23x27+27-23/24+23x27
A=23x27+4/24+23x27
A=4/24
A=1/6
A=\(\frac{23x27+27-23}{23x27+24}\)
A=\(\frac{23x27+4}{23x27+24}\)
A=\(\frac{625}{645}\)