tính bằng cách thuận tiện
\(4\cdot25\cdot0,25\cdot\dfrac{1}{5}\cdot\dfrac{1}{2}\cdot2=?\)
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`5/9xx1/4+4/9xx3/12`
`=5/9xx1/4+4/9xx1/4`
`=1/4xx(5/9+4/9)`
`=1/4xx9/9`
`=1/4xx1`
`=1/4`
Theo đề bài ta có:
\(B=\dfrac{-1^2.-2^2.....-100^2}{1.2.2.3.....99.100}\)
\(B=\dfrac{1^2.2^2.....100^2}{1.2.2.3.....99.100}\)
\(B=\dfrac{1.1.2.2......100.100}{1.2.2.3.....99.100}\)
\(B=\dfrac{1.2.3......100}{1.2.3.......99}.\dfrac{1.2.3......100}{2.3.4......100}\)
\(B=100\)
\(\dfrac{1}{7}.2\dfrac{1}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\dfrac{7}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\left(\dfrac{7}{3}-\dfrac{59}{6}\right)+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{1}{7}.\dfrac{-15}{2}+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{-15}{14}+\dfrac{15}{14}\)
= 0
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(A=\dfrac{3^6.45^4-15^{13}.\left(\dfrac{1}{5}\right)^9}{27^4.25^3+45^6}\)
\(=\dfrac{3^6.3^8.5^4-5^{13}.3^{13}.\left(\dfrac{1}{5}\right)^9}{3^{12}.5^6+3^{12}.5^6}\)
\(=\dfrac{3^{14}.5^4-5^{13}.3^{13}.\left(\dfrac{1}{5}\right)^9}{3^{12}.5^6+3^{12}.5^6}\)
\(=\dfrac{3^{13}\left[3.5^4-5^{13}.\left(\dfrac{1}{5}\right)^9\right]}{3^{12}(5^6+5^6)}\)
\(=\dfrac{3^{13}.1250}{3^{12}.31250}\)
\(=\dfrac{3^{13}.5^4.2}{3^{12}.5^6.2}\)
\(=\dfrac{3}{5^2}=\dfrac{3}{25}\)
Vậy \(A=\dfrac{3}{25}\)
4x25x0,25x1/5x1/2x2
=4x25x0,25x0,2x0,5x2
= (4x25)x(0,5x1)x(0,25x0,2)
= 100 x 1 x 0,05
= 100x0,05
= 5
`4.25.0,25. 1/5 . 1/2 . 2`
`=4.25. 1/4 . 1/5 . 1/2 . 2`
`=(4. 1/4).(25. 1/5).(1/2 .2)`
`=1.5.1=5`