35213 x 9258 + 94504 - 4625
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`B=(1+1/2).(1+1/3).(1+1/4)......(1+1/99)`
`B=(2/2+1/2).(3/3+1/3).(4/4+1/4).....(99/99+1/99)`
`B=3/2 . 4/3 . 5/4 ..... 100/99`
\(B=\dfrac{3.4.5......100}{2.3.4.5.....99}\)
`B=100/2`
`B=50`
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\(C=1-2+3-4+5-6+...+1011-1012+1013\\ =1+\left(3-2\right)+\left(5-4\right)+\left(7-6\right)+...+\left(1011-1010\right)+\left(1013-1012\right)\\ =1+1+1+1+...+1+1=507\)
(Từ `2` đến `1013` có `1012` số hạng => `506` nhóm, mỗi nhóm có giá trị là : `1` . Vậy tổng : `1+(3-2)+(5-4)+(7-6)+...+(1013-1012)=1+506=507`)
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\(600\) giây = \(\dfrac{600}{60}\) (phút) = \(10\) phút
\(\dfrac{1}{4}\) giờ = \(\dfrac{1}{4}\times60\) (phút) = \(15\) phút
\(\dfrac{3}{10}\) giờ = \(\dfrac{3}{10}\times60\) (phút) = \(18\) phút
Vì : `1` phút < `10` phút < `15` phút < `18` phút < `20` phút
=> Sắp xếp theo thứ tự tăng dần là : `1` phút ; `600` giây ; `1/4` giờ ; `3/10` giờ ; `20` phút
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\(\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\times...\times\left(1+\dfrac{1}{99}\right)\\ =\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\times...\times\left(\dfrac{99}{99}+\dfrac{1}{99}\right)\\ =\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{100}{99}\\ =\dfrac{3\times4\times5\times...\times100}{2\times3\times4\times...\times99}\\ =\dfrac{100}{2}=50\)
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\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+99\right)+\left(x+100\right)=5050\\ \left(x+x+x+...+x+x\right)+\left(1+2+3+...+99+100\right)=5050\\ 100\times x+\left(100+1\right)\times100:2=5050\\ 100\times x+101\times50=5050\\ 100\times x=5050-101\times50=0\\ x=0\)
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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 : \(2>\dfrac{2}{3}x3,5-\left(\dfrac{1}{10}+20\%\right):\dfrac{3}{5}\)
35213 x 9258 + 94504 - 4625 = 326091833
326091833