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\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{13}{4}+\dfrac{3}{4}-\dfrac{1}{20}=4+\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)=4+0=4\)
a ) \(\frac{25}{100}m^2>\frac{25}{100}ha=sai\)
b ) \(\frac{25}{100}m^2=\frac{25}{100}ha=sai\)
c ) \(\frac{25}{100}m^2< \frac{25}{100}ha=đúng\)
\(\frac{1}{4}\cdot45+0.25\cdot35+\frac{25}{100}\cdot19+25\%\)
\(=0.25\cdot45+0.25\cdot35+0.25\cdot19+0.25\)
\(=0.25\left(45+35+19+1\right)\)
\(=0.25\cdot100\)
\(=25\)
\(=0,25.45+0,25.35+0,25.19+0,25.1\)
\(=0,25.\left(45+35+19+1\right)\)
\(=0,25.100\)
\(=25\)
\(\dfrac{25}{100}\) + \(\dfrac{1}{4}\) x 29 + 25% \(\times\) 30 + 0,25 \(\times\) 40
= 0,25+ 0,25 x 29 + 0,25 x 30 + 0,25 x 40
= 0,25 x ( 1 + 29 + 30 + 40)
= 0,25 x 100
= 25
\(=0,25\times49+0,25+0,25\times31+0,25\times19\\ =0,25\times\left(49+1+31+19\right)\\ =0,25\times100\\ =25\)
Ta có:
\(A=\frac{1}{6.25}+\frac{1}{7.30}+...+\frac{1}{8.35}+\frac{1}{100.495}\)
\(=\frac{1}{6.\left(5.5\right)}+\frac{1}{7.\left(5.6\right)}+...+\frac{1}{8.\left(5.7\right)}+\frac{1}{100.\left(5.99\right)}\)
\(=\frac{1}{5}\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\right)\)
\(=\frac{1}{5}\left[\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\right]\)
\(=\frac{1}{5}\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)\)
Mà \(\frac{1}{5}-\frac{1}{100}< \frac{1}{5}\)nên \(A=\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)< \frac{1}{5}.\frac{1}{5}=\frac{1}{25}.\)
Vậy \(A< \frac{1}{25}.\)
100-5=95 phân số
(1/100+1/6):2=53/600
(495-25):5+1=95 số
(495+5)x95:2=23750
53/600x23750=25175/12
\(\dfrac{25}{100}\) + \(\dfrac{1}{4}\) x 401,7 + 25% x 304 + 0,25 x 293,3
= 0,25 x 1 + 0,25 x 401,7 + 0,25 x 304 + 0,25 x 293,3
= 0,25 x ( 1 + 401,7 + 304 + 293,3)
= 0,25 x 1000
= 250
= 0,25 x 1 + 0,25 x 401,7 + 0,25 x 304 + 0,25 x 293,3
= 0,25 x ( 1 + 401,7 + 304 + 293,3)
= 0,25 x 1000
= 250