Cm:(3525-3524)\(⋮\)17
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\(VT=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{100}\)
\(VP=\dfrac{1}{26}+\dfrac{1}{27}+...\dfrac{1}{50}=\dfrac{2}{26}+\dfrac{2}{28}+...\dfrac{2}{50}...\)
(VP lần lượt triển khai \(\dfrac{1}{26}=\dfrac{2}{26}-\dfrac{1}{26};\dfrac{1}{28}=\dfrac{2}{28}-\dfrac{1}{28}...\))
Tiếp tục \(\dfrac{2}{26}+\dfrac{2}{28}+...\dfrac{2}{50}...=\dfrac{1}{13}+\dfrac{1}{14}+...\dfrac{1}{25}...\)
(VP lần lượt triển khai \(\dfrac{1}{14}=\dfrac{2}{14}-\dfrac{1}{14};\dfrac{1}{16}=\dfrac{2}{16}-\dfrac{1}{16}...\))
Chuyển sang VT để đơn giản phần số đối \(-\dfrac{1}{2};\dfrac{1}{2}...\)
Cuối cùng ta sẽ được \(VT=1;VP=\dfrac{2}{2}\Rightarrow VT=VP\)
⇒Đpcm
\(-\dfrac{3}{16}+-\dfrac{3}{8}-\dfrac{5}{4}\)
\(=-\dfrac{3}{16}-\dfrac{3}{8}-\dfrac{5}{4}\)
\(=-\dfrac{3}{16}-\dfrac{6}{16}-\dfrac{20}{16}\)
\(=\dfrac{-3-6-20}{16}\)
\(=-\dfrac{29}{16}\)
\(\left(-\dfrac{25}{13}\right)+\dfrac{-9}{17}+\dfrac{12}{13}+\dfrac{25}{17}\)
\(=\left[\left(-\dfrac{25}{13}\right)+\dfrac{12}{13}\right]+\left[\left(-\dfrac{9}{17}+\dfrac{25}{17}\right)\right]\)
\(=-\dfrac{13}{13}+\dfrac{17}{17}\)
\(=-1+1\)
\(=0\)
\(-\dfrac{1}{12}+\left(-\dfrac{5}{6}\right)-\dfrac{4}{3}=-\dfrac{1}{12}-\dfrac{5}{6}-\dfrac{4}{3}=-\dfrac{1}{12}-\dfrac{10}{12}-\dfrac{16}{12}=-\dfrac{27}{12}=-\dfrac{9}{4}\)
\(6,5-\left(-\dfrac{1}{5}\right)=\dfrac{65}{10}+\dfrac{1}{5}=\dfrac{65}{10}+\dfrac{2}{10}=\dfrac{67}{10}\)
\(-\dfrac{18}{10}+0,4=-\dfrac{18}{10}+\dfrac{4}{10}=-\dfrac{14}{10}=-\dfrac{7}{5}\)
`@` `\text {Ans}`
`\downarrow`
`-1/8 - 3/20`
`= -5/40 - 6/40`
`= -11/40`
\(-\dfrac{1}{8}-\dfrac{3}{20}=-\dfrac{5}{40}-\dfrac{6}{40}=-\dfrac{11}{40}\)
\(\dfrac{-1}{16}+\dfrac{-1}{24}=\dfrac{-3}{48}+\dfrac{-2}{48}=\dfrac{-5}{48}\)
4,5 .( - \(\dfrac{7}{5}\))
= (\(\dfrac{4,5}{5}\)). (-7)
= 0,9. (-7)
= - 6,3
- \(\dfrac{14}{20}\) + 0,6
= - 0,7 + 0,6
= - (0,7 - 0,6)
= - 0,1
Ta có: \(35^{25}-35^{24}\)
\(=35^{24}\cdot\left(35-1\right)\)
\(=35^{24}\cdot34\)
Mà: 34 ⋮ 17
\(\Rightarrow35^{24}\cdot34\) ⋮ 17
Hay \(35^{25}-35^{24}\) ⋮ 17 (đpcm)
\(\left(35^{25}-35^{24}\right)=\left(35.35^{24}-35^{24}\right)=34.35^{24}\)
mà \(34⋮17\)
\(\Rightarrow34.35^{24}⋮17\)
⇒đpcm