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\(S=\dfrac{3}{2}+\dfrac{7}{6}+\dfrac{13}{12}+...+\dfrac{9901}{9900}\)

\(=1+\dfrac{1}{2}+1+\dfrac{1}{6}+...+1+\dfrac{1}{9900}\)

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

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

\(=99+\left(1-\dfrac{1}{100}\right)=100-\dfrac{1}{100}=\dfrac{10000-1}{100}=\dfrac{9999}{100}\)

1 tháng 7

S = ( 1+\(\dfrac{1}{2}\) ) + ( 1 + \(\dfrac{1}{6}\) ) + .... + ( 1 + \(\dfrac{1}{9900}\) )

   = 9900 + ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ..... + \(\dfrac{1}{99.100}\) )

   = 9900 + ( 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ..... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\) )

   = 9900 + 1 - \(\dfrac{1}{100}\)

   = 9901 - \(\dfrac{1}{100}\)

16 tháng 4 2022

= -103/273

= 0

= -5/7

16 tháng 4 2022

5.\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{-4}{12}=-\dfrac{103}{273}\)

b.\(-\dfrac{5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-5-2}{21}+\dfrac{8}{24}=-\dfrac{7}{21}+\dfrac{8}{24}=-\dfrac{1}{3}+\dfrac{8}{24}=0\)

c.\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+-\dfrac{5}{7}=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)

22 tháng 4 2017

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)

Vậy...

1 tháng 4 2021

\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)

a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)

\(=\dfrac{2}{3}+\dfrac{2}{7}\)

\(=\dfrac{14}{21}+\dfrac{6}{21}\)

\(=\dfrac{20}{21}\)

27 tháng 4 2023

6/21-(−12/44)+10/14−(1/(−4))−18/33

 

=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11

 

=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11

 

=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4

 

=−3/11+7/7+1/4=−3/11+7/7+1/4

 

=43/44

 

 

23 tháng 4 2021

viết p/s như nào vậy bạn

23 tháng 4 2021

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8 tháng 5 2017

1.

A=\(\dfrac{3\left|x\right|+2}{\left|x\right|-5}=\dfrac{3\left|x\right|-15+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)}{\left|x\right|-5}+\dfrac{17}{\left|x-5\right|}=3+\dfrac{17}{\left|x\right|-5}\)

Để A \(\in\)Z thì \(\left|x\right|-5\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)

Ta có :

\(\left|x\right|-5=-17\Rightarrow\left|x\right|=-12\left(KTM\right)\)

\(\left|x\right|-5=-1\Rightarrow\left|x\right|=4\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(\left|x\right|-5=1\Rightarrow\left|x\right|=6\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

\(\left|x\right|-5=17\Rightarrow\left|x\right|=32\Rightarrow\left[{}\begin{matrix}x=32\\x=-32\end{matrix}\right.\)

Vậy để A \(\in\)Z thì x \(\in\) {-32;-6;-4;4;6;32}

9 tháng 5 2017

thank nha

a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)

a: \(=\dfrac{-6}{11}:\dfrac{3\cdot11}{4\cdot5}=\dfrac{-6}{11}\cdot\dfrac{20}{33}=\dfrac{-2}{11}\cdot\dfrac{20}{11}=\dfrac{-40}{121}\)

b: \(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

c: \(=\dfrac{13}{10}:\dfrac{-5}{13}=\dfrac{-169}{50}\)

a) \(\frac{1}{3}+\frac{3}{4}-\frac{5}{6}\)

\(=\frac{1}{3}+\frac{3}{4}+\frac{-5}{6}\)

\(=\frac{1.4+3.3+\left(-5\right).2}{12}\)

\(=\frac{4+9+\left(-10\right)}{12}\)

\(=\frac{3}{12}\)

\(=\frac{1}{4}\)

b) \(\frac{-2}{3}+\frac{6}{5}\div\frac{2}{3}-\frac{2}{15}\text{ }\)

\(=\frac{-2}{3}+\frac{6}{5}\times\frac{3}{2}-\frac{2}{15}\)

\(=\frac{-2}{3}+\frac{18}{10}-\frac{2}{15}\)

\(=\frac{-2}{3}+\frac{9}{5}+\frac{-2}{15}\)

\(=\frac{\left(-2\right).5+9.3+\left(-2\right)}{15}\)

\(=\frac{\left(-10\right)+27+\left(-2\right)}{15}\)

\(=\frac{15}{15}\)

\(=1\)