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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
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: \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{210}\)
\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{420}\)
\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\right)\)
\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{20\cdot21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{21}\right)=2\cdot\dfrac{19}{42}=\dfrac{19}{21}\)
=(-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
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
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: \(\left(x;y\right)=\left(2;8\right);\left(7;3\right)\)