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a.
=\(\left(\dfrac{61}{7}+\dfrac{9}{7}\right)+\left(\dfrac{63}{20}+\dfrac{87}{20}\right)\)
=\(10+\dfrac{15}{2}\)
=\(\dfrac{35}{2}\)
b.
=\(\left(4\dfrac{5}{23}-3\dfrac{5}{23}\right)+\left(7\dfrac{7}{13}+6\dfrac{6}{13}\right)-2\dfrac{2}{5}\)
=\(1+14-\dfrac{12}{5}\)
=15-\(\dfrac{12}{5}\)
=\(\dfrac{63}{5}\)
Câu C khó khó mình chưa giải được !!!
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Ta đặt
\(A=\dfrac{7}{1\times2}+\dfrac{7}{2\times3}+...+\dfrac{7}{99\times100}\)
\(\dfrac{1}{7}\times A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{99\times100}\)
\(\dfrac{1}{7}\times A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\dfrac{1}{7}\times A=1-\dfrac{1}{100}\)
\(\dfrac{1}{7}\times A=\dfrac{99}{100}\)
\(A=\dfrac{99}{100}\div\dfrac{1}{7}\)
\(A=\dfrac{693}{100}\)
= 7.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100)
= 7.(1 - 1/100)
= 7 . 99/100
= 693/100
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Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$
$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$
$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương
$\Rightarrow x+2023=0$
$\Leftrightarrow x=-2023$
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\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\left(4\dfrac{3}{4}+\dfrac{1}{8}+3\dfrac{1}{12}\right)+\left[\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)\right]\)
\(=\left(\dfrac{19}{4}+\dfrac{1}{8}+\dfrac{37}{12}\right)+\left(-4,15\right)\)
\(=\dfrac{191}{24}-4,15\)
\(=\dfrac{457}{120}=3\dfrac{97}{120}\)
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\dfrac{19}{4}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{475}{100}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\dfrac{25}{200}+\left(-\dfrac{256}{200}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\left(-\dfrac{231}{200}\right)+\left(-\dfrac{30}{12}+\dfrac{37}{12}\right)\)
=
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\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
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\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)
\(\dfrac{3}{2}x-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\dfrac{3}{2}x=\dfrac{3}{5}+\dfrac{1}{5}\)
\(\dfrac{3}{2}x=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}:\dfrac{3}{2}\)
\(x=\dfrac{4}{5}\cdot\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{3}\)
\(x=\dfrac{9}{12}-\dfrac{4}{12}\)
\(x=\dfrac{5}{12}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x=\dfrac{1}{4}+\dfrac{2}{5}\)
\(\dfrac{3}{2}x=\dfrac{13}{20}\)
\(x=\dfrac{13}{20}:\dfrac{3}{2}\)
\(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)
\(x=\dfrac{13}{30}\)
\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{3}{8}\)
\(x=\dfrac{5}{4}\cdot\dfrac{8}{3}\)
\(x=\dfrac{10}{3}\)
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\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{13}-\dfrac{8}{15}}\)
\(=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}=\dfrac{1}{2}\)
`(x-3/8) +1/4= 7/8`
`=> x-3/8 = 7/8-1/4`
`=> x-3/8 = 7/8-2/8`
`=> x-3/8 =5/8`
`=>x= 5/8 +3/8`
`=>x= 8/8=1`