Question

[Nguyễn Chí Quân]

1. Tính nhanh

 a.A= 5x[1/5+1/7] -- [2/5+2/17+6/10+9/51]

 b.B=2003x14+1988+2001x2002/2002+2002x503+504x2002

 c.[4,58:3,27+5,232:3,27]x4,08--4,08 

 d.6/11+7/17+8/25+10/17+16/11+17/25

Answer 1

[Kiyotaka Ayanokoji]

Trả lời:

a, \(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{6}{10}+\frac{9}{51}\right)\)

\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)

\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{5}{5}+\frac{5}{17}\right)\)

\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-5\times\left(\frac{1}{5}+\frac{1}{17}\right)\)

\(A=5\times\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{5}-\frac{1}{7}\right)\)

\(A=5\times\left(\frac{1}{7}-\frac{1}{17}\right)\)

\(A=5\times\frac{10}{119}\)

\(A=\frac{50}{119}\)

  b, \(B=\frac{2003\times14+1988+2001\times2002}{2002+2002\times503+504\times2002}\)

\(B=\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)

\(B=\frac{2002\times14+14+1988+2001\times2002}{2002\times1008}\)

\(B=\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)

\(B=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}\)

\(B=\frac{2002\times2016}{2002\times1008}\)

\(B=2\)

c, Sửa dề 

\(\left(4,58\div3,27+5,23\div3,27\right)\times4,08-4,08\)

\(=\left[\left(4,58+5,23\right)\div3,27\right]\times4,08-4,08\)

\(=\left(9,81\div3,27\right)\times4,08-4,08\)

\(=3\times4,08-4,08\)

\(=4,08\times\left(3-1\right)\)

\(=4,08\times2\)

\(=8,16\)

d, \(\frac{6}{11}+\frac{7}{17}+\frac{8}{25}+\frac{10}{17}+\frac{16}{11}+\frac{17}{25}\)

\(=\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{17}+\frac{10}{17}\right)+\left(\frac{8}{25}+\frac{17}{25}\right)\)

\(=2+1+1\)

\(=4\)