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a: \(A=\dfrac{15}{4}-\dfrac{9}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
b: \(B=\dfrac{-3}{2}\cdot\dfrac{7-5\cdot4}{3}=\dfrac{-13}{-2}=\dfrac{13}{2}\)
c: \(C=\dfrac{-21}{10}\left(1-\dfrac{3}{4}\right)-\dfrac{3}{4}\)
\(=\dfrac{-21}{10}\cdot\dfrac{1}{4}-\dfrac{3}{4}\)
\(=\dfrac{-21}{40}-\dfrac{30}{40}=\dfrac{-51}{40}000000997\)
d: \(D=\dfrac{-2}{3}+\dfrac{-12}{5}-3=\dfrac{-10-36-45}{15}=\dfrac{-91}{15}\)
A = 2/1*5 + 2/5*9 + ... + 2/101*105
= 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)
= 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)
= 1/2(1 - 1/105)
= 1/2 * 104/105 = 52/105
Sửa câu b. Phân số thứ 2 phải là 4/5*8
B = 4/2*5 + 4/5*8 + ... + 4/47*50
= 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)
= 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)
= 4/3(1/2 - 1/50)
= 4/3 * 24/50 = 16/25
a) \(\frac{15}{16}\cdot\frac{4}{3}-\frac{1}{2}:\frac{5}{4}+3\)
\(=\frac{5}{4}-\frac{2}{5}+3\)
\(=\frac{25}{20}-\frac{8}{20}+\frac{60}{20}\)
\(=\frac{77}{20}=3\frac{17}{20}\)
b) \(\left(\frac{2}{3}-\frac{3}{8}\right):\left(\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{24}:\frac{17}{20}\)
\(=\frac{35}{102}\)
c) \(\frac{15}{8}\cdot\left(\frac{1}{3}+\frac{1}{8}\right)-\frac{3}{8}:\frac{3}{4}\)
\(=\frac{15}{8}\cdot\frac{11}{24}-\frac{1}{2}\)
\(=\frac{55}{64}-\frac{1}{2}\)
\(=\frac{23}{64}\)
d) \(\frac{20}{21}:\left(\frac{4}{5}-\frac{1}{10}\right)+\frac{13}{15}:\frac{5}{26}\)
\(=\frac{20}{21}:\frac{7}{10}+\frac{52}{15}\)
\(=\frac{200}{147}+\frac{52}{15}\)
\(=4\frac{608}{735}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
a. \(\dfrac{3}{4}\) : x - \(\dfrac{17}{5}\) = \(\dfrac{7}{10}\)
\(\dfrac{3}{4}\) : x = \(\dfrac{7}{10}\) + \(\dfrac{17}{5}\)
\(\dfrac{3}{4}\) : x = \(\dfrac{41}{10}\)
x = \(\dfrac{3}{4}\) : \(\dfrac{41}{10}\)
x = \(\dfrac{3}{4}\) . \(\dfrac{10}{41}\)
x = \(\dfrac{15}{82}\)
Vậy x = \(\dfrac{15}{82}\)
A . \(3\cdot\dfrac{5}{4}-\dfrac{3^2}{4}=\dfrac{15}{4}-\dfrac{9}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
B, \(\dfrac{-3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)=\dfrac{-3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{20}{3}\right)=\dfrac{-3}{2}\cdot\dfrac{-13}{2}=\dfrac{39}{4}\)
C , \(\dfrac{-21}{10}+\dfrac{21}{10}\cdot\dfrac{3}{4}-\dfrac{3}{4}=\dfrac{-84}{40}+\dfrac{63}{40}-\dfrac{30}{40}=\dfrac{-84+63-30}{40}=\dfrac{-51}{40}\)
D , \(\dfrac{-2}{3}+2\cdot\left(\dfrac{-2}{3}\right)\cdot3-3=\dfrac{-2}{3}\left(1+2\cdot3\right)-3=\dfrac{-2}{3}\cdot7-3=\dfrac{-14}{3}-\dfrac{9}{3}=-\dfrac{23}{3}\)