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a. \(\dfrac{-3}{12}+\dfrac{1}{-4}=\dfrac{-3}{12}+\dfrac{-3}{12}=\dfrac{-3-3}{12}=\dfrac{-6}{12}=\dfrac{-1}{2}\)
b. \(\dfrac{5}{12}+\dfrac{-3}{28}=\dfrac{35}{84}+\dfrac{-9}{84}=\dfrac{35+\left(-9\right)}{84}=\dfrac{26}{84}=\dfrac{13}{42}\)
c. \(\dfrac{-7}{15}+\dfrac{3}{35}=\dfrac{-49}{105}+\dfrac{9}{105}=\dfrac{-49+9}{105}=\dfrac{-40}{105}=\dfrac{-8}{21}\)
d. \(\dfrac{-5}{7}+\dfrac{-3}{4}=\dfrac{-20}{28}+\dfrac{-21}{28}=\dfrac{-20+\left(-21\right)}{28}=\dfrac{-41}{28}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
a: \(\dfrac{-1}{2}+\dfrac{2}{3}=\dfrac{-3+4}{6}=\dfrac{1}{6}\)
Số đối là -1/6
b \(-\dfrac{3}{4}+\dfrac{-4}{3}=\dfrac{-9-16}{12}=\dfrac{-25}{12}\)
Số đối là 25/12
c: \(\dfrac{-7}{2}+\dfrac{-3}{4}=\dfrac{-14-3}{4}=\dfrac{-17}{4}\)
Số đối là 17/4
d: \(-2-\dfrac{3}{4}=\dfrac{-8-3}{4}=-\dfrac{11}{4}\)
Số đối là 11/4
1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot10=14\)
2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
Ta có S = \(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+...+\dfrac{10}{2^{10}}\)
= \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\)
2S = 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)
2S - S = ( 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)) - ( \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\))
S = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}-\dfrac{10}{2^{10}}\)
Đặt A = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)
2A = 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)
2A - A = ( 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)) - ( 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\))
A = 2 - \(\dfrac{1}{2^9}\)
⇒ S = 2 - \(\dfrac{1}{2^9}\) - \(\dfrac{10}{2^{10}}\) = \(\dfrac{2^{11}}{2^{10}}-\dfrac{2}{2^{10}}-\dfrac{10}{2^{10}}=\dfrac{2^2\left(2^9-3\right)}{2^{10}}=\dfrac{2^9-3}{2^8}\)
Vậy S = \(\dfrac{2^9-3}{2^8}\)
a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)
\(=1-\dfrac{-5}{2}\)
\(=\dfrac{2}{2}-\dfrac{-5}{2}\)
\(=\dfrac{7}{2}\)
dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!
Giải:
a)(4-12/5).25/8-2/5:-4/25
=8/5.25/8-(-5/2)
=5+5/2
=15/2
b)(-5/24+3/4-7/12):(-5/16)
=-1/24:(-5/16)
=2/15
c)6/7+5/4:(-5)-(-1/28).(-2)2
=6/7+(-1/4)-(-1/28).4
=6/7-1/4-(-1/7)
=6/7-1/4+1/7
=(6/7+1/7)-1/4
=1-1/4
=3/4
Chúc bạn học tốt!
1) Ta có
\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)
\(C=\dfrac{1}{2022}\)
2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{29}{30}\)
\(=\dfrac{1.2.3...29}{2.3.4...30}\)
\(=\dfrac{1}{30}\)
\(B=1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.1\dfrac{1}{24}...1\dfrac{1}{168}\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{169}{168}\)
\(=\dfrac{4.9.16.25...169}{3.8.15.24...168}\)
\(=\dfrac{2.3.4...13}{1.2.3...12}.\dfrac{2.3.4...13}{3.4.5...14}\)
\(=13.\dfrac{1}{7}\)
\(=\dfrac{13}{7}\).
A = \(\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{29}{30}=1x1x1x...x\dfrac{1}{30}=\dfrac{1}{30}\)
B = \(\dfrac{4}{3}x\dfrac{9}{8}x\dfrac{16}{15}x\dfrac{25}{24}x...x\dfrac{169}{168}=1x1x1x1x...x\dfrac{13}{7}=\dfrac{13}{7}\)
Câu B em chưa rõ cách làm nhanh cho lắm. Nếu ko cần tính nhanh thì chị có thể giải bình thường ra giấy ha.
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
4/28 + 4/168 + 4/408 + 4/748 + 4/1188 + 4/1728
= 1/7 + 1/42 + 1/102 + 1/187 + 1/297 + 1/432
= 1/7 + 1/7.6 + 1/6.17 + 1/17.11 + 1/11.27 + 1/27.16
= 1/7(1 + 1/6) + 1/17(1/6 + 1/11) + 1/27(1/11 + 1/16)
= 1/7.7/6 + 1/17.17/66 + 1/27.27/176
= 1/6 + 1/66 + 1/176
= 12/66 + 1/176
= 3/16