2 và 2/3x 3,5-(1/10+20%):3/5
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28/15 : 75/100 - ( 3/5 + 25/100) : 3/4 - 51/13 : 3
= 112/45 - 17/20 : 3/4 - 17/13
= 112/45 - 17/15 - 17/13
= 61/45 - 17/13
= 28/585
28/15 : 75/100 - ( 3/5 + 25/100) : 3/4 - 51/13: 3
= \(\dfrac{112}{45}\) - \(\dfrac{17}{20}\): \(\dfrac{3}{4}\) - \(\dfrac{17}{13}\)
= \(\dfrac{112}{45}\) - \(\dfrac{17}{15}\) - \(\dfrac{17}{13}\)
= \(\dfrac{1456}{585}\) - \(\dfrac{663}{585}\) - \(\dfrac{765}{585}\)
= \(\dfrac{28}{585}\)
113/15 : 0,75 - ( 3/5 + 25%) : 3,4 - 312/13 : 3
= 1 : 0,75 - (3/5 + 1/4): 3,4 - 312/13 : 3
= 100/75 - 17/20 : 3,4 - 3-1/13
= 4/3 - 17/20 : 34/10 - 3-1/13
= 4/3 - 1/4 - 3-1/13
= 13/12 - 3-1/13
Sửa tử phân số cuối là `15`
\(\dfrac{15}{90 \times 94}+\dfrac{15}{94 \times 98}+...+\dfrac{15}{146 \times 150}\)
\(=\dfrac{15}{4} \times (\dfrac{4}{90 \times 94}+\dfrac{4}{94 \times 98}+...+\dfrac{4}{146 \times 150})\)
\(=\dfrac{15}{4} \times (\dfrac{1}{90}-\dfrac{1}{94}+\dfrac{1}{94}-\dfrac{1}{98}+...+\dfrac{1}{146}-\dfrac{1}{150})\)
\(=\dfrac{15}{4} \times (\dfrac{1}{90}-\dfrac{1}{150})\)
\(=\dfrac{15}{4} \times \dfrac{1}{225}\)
\(=\dfrac{1}{60}\)
32: ( 23. \(\dfrac{1}{16}\))
= 32 : (8.\(\dfrac{1}{16}\))
= 32 : \(\dfrac{1}{2}\)
= 64
52.35. (\(\dfrac{3}{5}\))2
= 52.35. \(\dfrac{3^2}{5^2}\)
= 37
=2187
(\(\dfrac{1}{7}\))2.\(\dfrac{1}{7}\).492
= \(\dfrac{1}{7^3}\). 74
= 7
`a)a^{3}.a^{5}=a^{3+5}=a^8`
`b)x^{7}.x.x^{4}=x^{7+1+4}=x^{12}`
`c)3^{5}.4^{5}=(3.4)^{5}=12^5`
`d)8^{5}.2^{3}=(2^{3})^{5}.2^{3}=2^{15}.2^{3}=2^{15+3}=2^{18}`
2 và 2/3x 3,5-(1/10+20%):3/5
2/3x 3,5-(1/10+20%):3/5
= 2/3 x 35/10 - ( 1/10 + 20/100 ) : 3/5
= 2/3 x 7/2 - ( 1/10 - 1/5) : 3/5
= 7/3 - ( -1/10) : 3/5
= 73/30 : 3/5
= 73/18 \(\approx\) 4.05
=> 2 < 4,05
\(\dfrac{2}{3}\times3,5-\left(\dfrac{1}{10}+20\%\right):\dfrac{3}{5}\\ =\dfrac{2}{3}\times\dfrac{7}{2}-\left(\dfrac{1}{10}+\dfrac{20}{100}\right).\dfrac{5}{3}\\ =\dfrac{7}{3}-\left(\dfrac{1}{10}+\dfrac{2}{10}\right).\dfrac{5}{3}\\ =\dfrac{7}{3}-\dfrac{3}{10}.\dfrac{5}{3}\\ =\dfrac{7}{3}-\dfrac{5}{10}\\ =\dfrac{7}{3}-\dfrac{1}{2}\\ =\dfrac{14}{6}-\dfrac{3}{6}=\dfrac{11}{6}\)
\(2=\dfrac{12}{6}>\dfrac{11}{6}\)
Vậy : \(2>\dfrac{2}{3}x3,5-\left(\dfrac{1}{10}+20\%\right):\dfrac{3}{5}\)