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\(A=3\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+.....+\frac{3}{55\cdot58}\right)\)
\(A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{55}-\frac{1}{58}\right)\)
\(A=3\left(1-\frac{1}{58}\right)\)
\(A=3-\frac{1}{174}< 3< \frac{10}{3}\)
Ta có :
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}\)
\(S=\frac{2^{10}.3-3}{2^9}\)
Vậy \(S=\frac{2^{10}.3-3}{2^9}\)
vận dụng 3S lên
xong tìm S nha bn ok
tại k có thời gian nên chỉ giúp thế thôi
a,8/3x + 26/3= 10/3
8/3x = 10/3- 26/3 = -16/3
=>x = -16/3 : 8/3 = -2
b, (2/3-1/2)x = 5/12
1/6x = 5/12
=>x = 5/2
xong rùi đó
nhớ tk nha
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)
\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)
\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)
\(=\frac{1}{30}\cdot\frac{31}{2}\)
\(=\frac{31}{60}\)
b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)
\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)
Mà \(\frac{3}{2}< 2\)
\(\Rightarrow1< A< 2\)
c ,Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
NHân 3S nhé e
Ta có:
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\)
=> \(2S=6+3+\frac{3}{2}+...+\frac{3}{2^9}\)
=> \(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^9}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\right)\)
=> \(S=6-\frac{3}{2^{10}}=6-\frac{3}{1024}=\frac{6144}{1024}-\frac{3}{1024}=\frac{6141}{1024}\)(Vì \(2S-S=S\))
T**k mik nhé!;
Hok tốt!