Question

Bài 1: Tính

\(\dfrac{1}{4}+\dfrac{5}{10}-\dfrac{1}{3}-\dfrac{7}{8}\)

Bài 2:Tính bằng cách hợp lí

a) A= \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

b) B= \(\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

c) C= \(\dfrac{-5}{7}.\:\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

d) D= \(0,7.\:2\dfrac{2}{3}.20.\:0,375.\:\dfrac{5}{28}\)

e) E= \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

Bài 5: Tính nhanh

A=\(25.20,04+4.20,04-2004.20,33+2004.20,04\)

Bài 6: So sánh

\(\dfrac{10^5+4}{10^5-1}\)và \(\dfrac{10^5+3}{10^5-2}\)

Answer 1

Bài 2: 

a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)

\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)

b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)

\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)

c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)

d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)