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15 tháng 10 2023

m) Ta có số: \(\overline{24x2y}\) số này chia 5 dư 2

Mà số chia hết cho 5 có chữ số tận cùng là 0 hoặc 5 nên khi số đó chia 5 dư 2 thì số đó phải có chữ số tận cùng là 2 hoặc 7 

\(\Rightarrow y\in\left\{2;7\right\}\)

Số này lại chia hết cho 9 nên \(2+4+x+2+y=8+x+y\) ⋮ 9

Với \(y=2\)

\(\Rightarrow8+x+2=18\)

\(\Rightarrow x=18-10=8\)

Với \(y=7\)

\(\Rightarrow8+x+7=18\)

\(\Rightarrow x=18-15=3\) 

Vậy: \(\left(x;y\right)=\left(2;8\right);\left(7;3\right)\)

m: \(\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)

\(=\dfrac{2}{4}+\dfrac{2}{28}+\dfrac{2}{70}+\dfrac{2}{130}+\dfrac{2}{208}+\dfrac{2}{304}\)

\(=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{10\cdot13}+\dfrac{2}{13\cdot16}+\dfrac{2}{16\cdot19}\)

\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{16}-\dfrac{1}{19}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{18}{19}=\dfrac{12}{19}\)

e: \(\dfrac{4}{5}+\dfrac{1}{5}\left(x+1\right)=-\dfrac{2}{5}\)

=>\(\dfrac{1}{5}\left(x+1\right)=-\dfrac{2}{5}-\dfrac{4}{5}=-\dfrac{6}{5}\)

=>x+1=-6

=>x=-6-1=-7

g: \(\dfrac{5}{4}-\dfrac{1}{4}:\left(-\dfrac{1}{2}+x\right)=-1\)

=>\(\dfrac{1}{4}:\left(x-\dfrac{1}{2}\right)=\dfrac{5}{4}+1=\dfrac{9}{4}\)

=>\(x-\dfrac{1}{2}=\dfrac{1}{4}:\dfrac{9}{4}=\dfrac{1}{9}\)

=>\(x=\dfrac{1}{9}+\dfrac{1}{2}=\dfrac{11}{18}\)

p: \(\left(-\dfrac{2}{5}\right)^2+\dfrac{17}{-18}\cdot\dfrac{36}{34}-\left(-\dfrac{2}{3}\right)^3\)

\(=\dfrac{4}{25}-\dfrac{17}{34}\cdot\dfrac{36}{18}-\dfrac{-8}{27}\)

\(=\dfrac{4}{25}+\dfrac{8}{27}-1=\dfrac{-367}{675}\)

q: \(\left(-\dfrac{1}{2}\right)^0-\dfrac{-1}{3}\cdot\dfrac{-9}{12}+\dfrac{2^4}{-4}\)

\(=1-\dfrac{1}{3}\cdot\dfrac{3}{4}+\dfrac{16}{-4}\)

\(=1-\dfrac{1}{4}-4=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)

r: \(\left(-5\right)\cdot\dfrac{17}{45}-\left(-\dfrac{2}{3}\right)^2+\left(-\dfrac{20}{2023}\right)^0\)

\(=-\dfrac{17}{9}-\dfrac{4}{9}+1\)

\(=-\dfrac{21}{9}+1=-\dfrac{12}{9}=-\dfrac{4}{3}\)

z4:

\(\dfrac{24}{148}=\dfrac{6}{37}=\dfrac{108}{37\cdot18}\)

\(\dfrac{-14}{-36}=\dfrac{7}{18}=\dfrac{7\cdot37}{18\cdot37}=\dfrac{259}{37\cdot18}\)

mà 108<259

nên \(\dfrac{24}{148}< \dfrac{-14}{-36}\)

z5: \(\dfrac{-26}{-72}=\dfrac{26}{72}< 1\)

\(1< \dfrac{45}{20}=\dfrac{-45}{-20}\)

Do đó: \(\dfrac{-26}{-72}< \dfrac{-45}{-20}\)

z6: \(\dfrac{14}{42}=\dfrac{1}{3}=\dfrac{1\cdot4}{3\cdot4}=\dfrac{4}{12}\)

\(\dfrac{21}{28}=\dfrac{3}{4}=\dfrac{3\cdot3}{4\cdot3}=\dfrac{9}{12}\)

mà 4<9

nên \(\dfrac{14}{42}< \dfrac{21}{28}\)

z7: \(\dfrac{-14}{-56}=\dfrac{1}{4}=\dfrac{5}{20}\)

\(\dfrac{21}{35}=\dfrac{3}{5}=\dfrac{3\cdot4}{5\cdot4}=\dfrac{12}{20}\)

mà 5<12

nên \(\dfrac{-14}{-56}< \dfrac{21}{35}\)

z8: \(10A=\dfrac{10^{201}+10}{10^{201}+1}=1+\dfrac{9}{10^{201}+1}\)

\(10B=\dfrac{10^{202}+10}{10^{202}+1}=1+\dfrac{9}{10^{202}+1}\)

\(10^{201}+1< 10^{202}+1\)

=>\(\dfrac{9}{10^{201}+1}>\dfrac{9}{10^{202}+1}\)

=>\(\dfrac{9}{10^{201}+1}+1>\dfrac{9}{10^{202}+1}+1\)

=>10A>10B

=>A>B

 

 

14 tháng 3 2022

B

A

A

A

14 tháng 3 2022

p: \(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+...+\dfrac{5}{50\cdot51}\)

\(=5\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{50\cdot51}\right)\)

\(=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)

\(=5\cdot\left(1-\dfrac{1}{51}\right)=5\cdot\dfrac{50}{51}=\dfrac{250}{51}\)

q: \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{210}\)

\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{420}\)

\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\right)\)

\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{20\cdot21}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{21}\right)=2\cdot\dfrac{19}{42}=\dfrac{19}{21}\)

26 tháng 7 2022

=(-1+2)-(3+4)-(5+6)-........-(2017+2018)

=1-7-11-........-4035

=-1009

 

A=1/2^2+1/3^2+...+1/23^2

=>A<1-1/2+1/2-1/3+...+1/22-1/23

=>A<22/23

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)