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a: =27/45-20/45=7/45
b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)
c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)
a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)
b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)
c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)
d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)
Ta có: \(B=\dfrac{\dfrac{1}{22}-\dfrac{1}{2}+\dfrac{1}{13}}{\dfrac{3}{22}-\dfrac{3}{2}+\dfrac{3}{13}}\cdot\dfrac{\dfrac{3}{4}-0.375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0.875}+\dfrac{3}{4}\)
\(=\dfrac{1}{3}\cdot\dfrac{-15}{4}+\dfrac{3}{4}\)
\(=\dfrac{-5}{4}+\dfrac{3}{4}=\dfrac{-1}{2}\)
`@` `\text {Ans}`
`\downarrow`
`j)`
`-3/4 + 2/7 + (-1)/4 + 3/5 + 5/7`
`= (-3/4 - 1/4) + (2/7 + 5/7) + 3/5`
`= -1 + 1 + 3/5`
`= 3/5`
`k)`
`-2/17 + 15/23 + (-15)/17 + 4/19 + 8/23`
`= (-2/17 - 15/17) + (15/23 + 8/23) + 4/19`
`= -1 + 1 + 4/19`
`= 4/19`
$#KDN040510$
j: =-3/4-1/4+2/7+5/7+3/5
=-1+1+3/5
=3/5
k: =-2/17-15/17+15/23+8/23+4/19
=-1+1+4/19
=4/19
\(B=\dfrac{\dfrac{2}{3}\left(\dfrac{17}{3}-\dfrac{4}{13}\right)+\dfrac{7}{3}}{\dfrac{9}{20}}=3:\dfrac{9}{20}=3\cdot\dfrac{20}{9}=\dfrac{60}{9}=\dfrac{20}{3}\)
\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
\(K=\dfrac{9-5}{3}+\dfrac{2.9-5}{3^2}+\dfrac{3.9-5}{3^3}+...+\dfrac{101.9-5}{3^{101}}\)
\(K=\dfrac{9}{3}+\dfrac{2.9}{3^2}+\dfrac{3.9}{3^3}+...+\dfrac{101.9}{3^{101}}-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(K=9\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{101}{3^{101}}\right)-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(K=9A-5B\)
Xét \(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{101}{3^{101}}\) (1)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{2}{3^3}+\dfrac{3}{3^4}+...+\dfrac{100}{3^{101}}+\dfrac{101}{3^{102}}\) (2)
Trừ vế với vế (1) cho (2):
\(\dfrac{2}{3}A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}-\dfrac{101}{3^{102}}=B-\dfrac{101}{3^{102}}\)
\(\Rightarrow A=\dfrac{3}{2}\left(B-\dfrac{101}{3^{102}}\right)\Rightarrow K=\dfrac{27}{2}\left(B-\dfrac{101}{3^{102}}\right)-5B\)
\(\Rightarrow K=\dfrac{17}{2}B-\dfrac{27}{2}.\dfrac{101}{3^{102}}\)
Xét \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{90}}+\dfrac{1}{3^{100}}\)
\(\Rightarrow3B-1+\dfrac{1}{3^{101}}=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}=B\)
\(\Rightarrow2B=1-\dfrac{1}{3^{101}}\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{101}}\)
\(\Rightarrow K=\dfrac{17}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{101}}\right)-\dfrac{27}{2}.\dfrac{101}{3^{102}}\)
\(\Rightarrow K=\dfrac{17}{4}-\dfrac{1}{3^{101}}\left(\dfrac{17}{4}+\dfrac{27.101}{6}\right)< \dfrac{17}{4}\) (đpcm)
good job