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Xin lỗi mình viết nhầm
A = 2 + 22 + 23 + 24 + ... + 2199 + 2200
A = ( 2 + 22 + 23 + 24 ) + ... + ( 2197 + 2198 + 2199 + 2200 )
A = 2 . ( 1 + 2 + 22 + 23 ) + ... + 2197 ( 1 + 2 + 22 + 23 )
A = 2 . 15 + ... + 2197 . 15
A = ( 2 + ... + 2197 ) .15 \(\Rightarrow A⋮15\)
Ta có:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right).\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\)
\(\Rightarrow C^2< \frac{1}{201}\left(dpcm\right)\)
Hình như mik chưa tính nhưng vế sau là = 0 nên bnj ko cần tính vế trước đâu
( \(\frac{12}{199}\)+ \(\frac{23}{200}\)- \(\frac{34}{201}\)) x ( \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{6}\))
= ( \(\frac{12}{199}\)+ \(\frac{23}{200}\)- \(\frac{34}{201}\)) x ( \(\frac{3}{6}\)- \(\frac{2}{6}\)- \(\frac{1}{6}\))
=( \(\frac{12}{199}\) + \(\frac{23}{200}\) - \(\frac{34}{201}\)) x 0
= 0
Học tốt ^-^
vì\(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=\frac{3}{6}-\frac{2}{6}-\frac{1}{6}=\frac{3-2-1}{6}=\frac{0}{6}=0\)
=> \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0=0\)
a) \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\)\(\times\)\(\dfrac{21}{39}+\dfrac{49}{91}\)\(\times\)\(\dfrac{8}{15}\)
= \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\times\dfrac{7}{13}+\dfrac{7}{13}\times\dfrac{8}{15}\)
= \(\dfrac{7}{13}\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)
= \(\dfrac{7}{13}\times\dfrac{7}{12}\)
= \(\dfrac{49}{156}\)
b) \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
= \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times0\)
= 0
a: \(=\dfrac{1}{3}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}+\dfrac{-5}{2}+\dfrac{3}{2}\)
\(=\dfrac{1}{3}+\dfrac{1}{8}-1-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{8+3-24}{24}-\dfrac{165}{100}\)
\(=\dfrac{-263}{120}\)
b: \(=\dfrac{5}{22}+\dfrac{3}{13}-\dfrac{13}{8}-\dfrac{2}{11}+\dfrac{3}{2}\)
\(=\dfrac{5}{22}-\dfrac{4}{22}-\dfrac{13}{8}+\dfrac{12}{8}+\dfrac{3}{13}\)
\(=\dfrac{1}{22}-\dfrac{1}{8}+\dfrac{3}{13}=\dfrac{173}{1144}\)
a: \(=\dfrac{1}{3}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)
\(=\dfrac{11}{24}+\left(-\dfrac{37}{100}+\dfrac{1}{8}-1\right)\)
\(=\dfrac{11}{24}-\dfrac{249}{200}=-\dfrac{59}{75}\)
b: \(=\dfrac{5}{22}+\dfrac{3}{13}-\dfrac{13}{8}-\dfrac{2}{11}+\dfrac{3}{2}\)
\(=\dfrac{5}{22}-\dfrac{4}{22}+\dfrac{3}{13}-\dfrac{13}{8}+\dfrac{12}{8}\)
\(=\dfrac{1}{22}+\dfrac{3}{13}-\dfrac{1}{8}=\dfrac{173}{1144}\)