` sqrt/ac

số tự nhiên có bao nhiêu chữ số ạ?

`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`

=)

\(C=1-2+3-4+5-6+...+1011-1012+1013\\ =1+\left(3-2\right)+\left(5-4\right)+\left(7-6\right)+...+\left(1011-1010\right)+\left(1013-1012\right)\\ =1+1+1+1+...+1+1=507\)

(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`)

a)\(5^{x+2}=125=5^3\\ =>x+2=3\\ =>x=1\)

c)\(5^{2x-1}-2.5^2=3.5^2\\ =>5^{2x-1}=3.5^2+2.5^2\\ =>5^{2x-1}=5^2.\left(3+2\right)=5^2.5=5^3\\ =>2x-1=3\\ =>x=2\)

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 : \(2>\dfrac{2}{3}x3,5-\left(\dfrac{1}{10}+20\%\right):\dfrac{3}{5}\)

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

= \(\dfrac{112}{45}\) - \(\dfrac{17}{20}\): \(\dfrac{3}{4}\) - \(\dfrac{17}{13}\)

= \(\dfrac{112}{45}\) - \(\dfrac{17}{15}\) - \(\dfrac{17}{13}\)

= \(\dfrac{1456}{585}\) - \(\dfrac{663}{585}\) - \(\dfrac{765}{585}\)

= \(\dfrac{28}{585}\)

  1^13/15 : 0,75 - ( 3/5 + 25%) : 3,4 - 3^12/13 : 3

= 1 : 0,75 - (3/5 + 1/4): 3,4 - 3^12/13 : 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