Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
\(14.a)A=\dfrac{-3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\)
\(A=\dfrac{-3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}+\dfrac{-3}{5}+\dfrac{9}{17}\)
\(A=\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)+\dfrac{-3}{5}\)
\(A=\left(-1\right)+1+\dfrac{-3}{5}\)
\(A=0+\dfrac{-3}{5}=\dfrac{-3}{5}\)
\(b)B=\dfrac{7}{5}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right)\div\dfrac{5}{8}\)
\(B=\dfrac{7}{5}.\dfrac{-15}{14}+\dfrac{25}{16}\div\dfrac{5}{8}\)
\(B=\left(\dfrac{-3}{2}\right)+\dfrac{5}{2}=1\)
\(c)C=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(C=1-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-...-\dfrac{2}{99}+\dfrac{2}{99}-\dfrac{2}{100}\)
\(C=1-\dfrac{2}{100}=1-\dfrac{1}{50}=\dfrac{49}{50}\)
\(15.a)2x+\dfrac{-1}{4}=\dfrac{3}{2}\)
\(2x\) \(=\dfrac{3}{2}-\left(\dfrac{-1}{4}\right)=\dfrac{3}{2}+\dfrac{1}{4}=\dfrac{7}{4}\)
\(x\) \(=\dfrac{7}{4}\div2=\dfrac{7}{8}\)
\(b)\dfrac{15}{x}=\dfrac{-3}{4}\)
\(\Rightarrow x=\dfrac{15.4}{-3}=-20\)
