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\(=\frac{3}{4}\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{41.45}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)\)
\(=\frac{3}{4}\times\frac{8}{45}\)
\(=\frac{2}{15}\)
\(A=\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+...+\dfrac{1}{87\times89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{87}-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{87}-\dfrac{1}{87}\right)-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{89}\)
\(A=\dfrac{84}{445}\)
Vậy, `A=84/445.`
A = \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+\(\dfrac{1}{9\times11}\)+...+\(\dfrac{1}{87\times89}\)
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{2}{5\times7}\)+\(\dfrac{2}{7\times9}\)+\(\dfrac{2}{9\times11}\)+...+\(\dfrac{2}{87\times89}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) +...+ \(\dfrac{1}{87}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{84}{445}\)
A = \(\dfrac{42}{445}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)
\(45\cdot\left(-\frac{5}{7}\right)+9\cdot5+\frac{8}{9}+135+45\cdot\frac{52}{63}\)
\(=-\frac{225}{7}+45+\frac{8}{9}+135+\frac{260}{7}\)
\(=\left(-\frac{225}{7}+\frac{260}{7}\right)+\left(45+135\right)+\frac{8}{9}\)
\(=\frac{35}{7}+180+\frac{8}{9}\)
\(=5+180+\frac{8}{9}\)
\(=185+\frac{8}{9}=185\frac{8}{9}\)
tick đúng cho tớ nha
a)<=>2/7:(13/24+5/24) b)<=>15/24-9/24-4/24 c)=7.5.39/13.14.15
<=>2/7:3/4 <=>2/24=1/12 =3/2.3
<=>2/7x4/3 =1/2
<=>8/21
a. \(\frac{2}{7}:\frac{13}{24}+\frac{2}{7}:\frac{5}{24}\)
= \(\frac{2}{7}:\left(\frac{13}{24}+\frac{5}{24}\right)\)
= \(\frac{2}{7}:\frac{3}{4}\)
= \(\frac{8}{21}\)
b. \(\frac{15}{24}-\frac{3}{8}-\frac{1}{6}\)
= \(\frac{15}{24}-\frac{9}{24}-\frac{4}{24}\)
= \(\frac{2}{24}=\frac{1}{12}\)
c. \(\frac{7}{13}.\frac{5}{14}.\frac{39}{15}\)
= \(\frac{7.5.3.13}{13.2.7.3.5}\)
= \(\frac{1}{2}\)
\(3A=\frac{6}{3\times\left(3+6\right)}+\frac{15}{9\times\left(9+15\right)}+...+\frac{39}{84\times\left(84+39\right)}\)
\(=\frac{1}{3}-\frac{1}{9}+\frac{1}{9}-\frac{1}{24}+...+\frac{1}{84}-\frac{1}{123}=\frac{1}{3}-\frac{1}{123}=\frac{40}{123}\)
\(\Rightarrow A=\frac{40}{3.123}=\frac{40}{369}\)
mình biết mà tính mệt quá