Question

[Nguyễn Xuân Mai]

 

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 9:

a: \(A=\left(\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}\right)+\left(\dfrac{2}{3}+\dfrac{-7}{15}+1\dfrac{1}{7}\right)\)

\(=\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}+\dfrac{2}{3}+\dfrac{-7}{15}+\dfrac{8}{7}\)

\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(-\dfrac{8}{15}-\dfrac{7}{15}\right)+\left(-\dfrac{1}{7}+\dfrac{8}{7}\right)\)

\(=1-1+1=1\)

b: \(B=0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

\(=0,25+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1,25\)

\(=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(0,25-1,25\right)-\dfrac{1}{8}\)

\(=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)

c: \(C=\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}+\dfrac{6}{7}-\dfrac{5}{6}+\dfrac{4}{5}-\dfrac{3}{4}\)

\(=\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{4}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{6}-\dfrac{5}{6}\right)+\left(-\dfrac{6}{7}+\dfrac{6}{7}\right)+\dfrac{7}{8}\)

\(=0+0+0+0+\dfrac{7}{8}=\dfrac{7}{8}\)

d: \(D=\left(2025-\dfrac{5}{181}+\dfrac{1}{50}\right)-\left(4+\dfrac{3}{181}-\dfrac{2}{50}\right)-\left(1-\dfrac{8}{181}+\dfrac{3}{50}\right)\)

\(=2025-\dfrac{5}{181}+\dfrac{1}{50}-4-\dfrac{3}{181}+\dfrac{2}{50}-1+\dfrac{8}{181}-\dfrac{3}{50}\)

\(=2025-4-1=2020\)

Answer 2

[Nguyễn Xuân Mai]

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