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31 tháng 10 2021

Mình nhầm ạ môn kia là Toán ạ

31 tháng 10 2021

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25 tháng 1 2021

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25 tháng 1 2021

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8 tháng 8 2018

c. Có \(\overline{ab}+\overline{ba}=10a+b+10b+a\)

\(=\left(10a+a\right)+\left(10b+b\right)\)

\(=11a+11b\)

\(=11.\left(a+b\right)\)

Ta thấy \(11.\left(a+b\right)⋮11\)

Vậy \(\overline{ab}+\overline{ba}⋮11\left(dpcm\right)\)

a: \(5C=5+5^2+5^3+...+5^{2018}\)

\(\Leftrightarrow4C=5^{2018}-1\)

\(\Leftrightarrow C=\dfrac{5^{2018}-1}{4}\)

\(\Leftrightarrow5^x-1=\dfrac{5^{2018}-1}{4}\)

\(\Leftrightarrow5^x=\dfrac{5^{2018}+3}{4}\)(vô lý)

c: \(64^{10}-32^{11}-16^{13}\)

\(=2^{60}-2^{55}-2^{52}\)

\(=2^{52}\left(2^8-2^3-1\right)\)

\(=2^{52}\cdot247⋮̸49\)

20 tháng 7 2018

các bạn giải giùm mình nha ,nếu giúp được thì các bạn sẽ giải hết bài cho mình nha.bạn nào làm được hết mình sẽ tích đúng cho luôn.mình cần bài này trong 3 tiếng nữa .

16 tháng 2 2020

ban chia ra tung bai di dai lam

16 tháng 2 2020

bai nao lam dc thi giam di nhe

23 tháng 1 2020

                                                                Bài giải

a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)

\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)

b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)

\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)

c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)

d,  \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)

\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)

\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)

\(=1+1+1\)

\(=3\)

e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)

\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)

\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)

\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)

\(=1+1+1\)

\(=3\)

23 tháng 1 2020

a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)

\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)

b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)

\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)

c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)

d,  \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)

\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)

\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)

\(=1+1+1\)

\(=3\)

e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)

\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)

\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)

\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)

\(=1+1+1\)

\(=3\)

a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)

\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)

=>x+1=0

hay x=-1

c: |x-2|=13

=>x-2=13 hoặc x-2=-13

=>x=15 hoặc x=-11

d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)

=>7|x-2|=5/3

=>|x-2|=5/21

=>x-2=5/21 hoặc x-2=-5/21

=>x=47/21 hoặc x=37/21