A = 4\(\dfrac{1}{3}\) + 3\(\dfrac{3}{4}\) + 2 - \(\dfrac{2}{5}\) + 8 - \(\dfrac{3}{5}\) + 7 - \(\dfrac{1}{4}\) + 6 - \(\dfrac{2}{3}\)

A = \(\dfrac{13}{3}\) + \(\dfrac{15}{4}\) + ( 2 + 8 + 6 + 7) - ( \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) - \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\)

A = \(\dfrac{13}{3}\) + \(\dfrac{15}{4}\) + 23 - 1 - \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\)

A = ( \(\dfrac{13}{3}\) - \(\dfrac{2}{3}\)) + 22 +( \(\dfrac{15}{4}\) - \(\dfrac{1}{4}\) )

A = \(\dfrac{11}{3}\) + 22 + \(\dfrac{14}{4}\)

A =  \(\dfrac{11}{3}\) + 22 + \(\dfrac{7}{2}\)

A =    \(\dfrac{22}{6}\) + \(\dfrac{132}{6}\) + \(\dfrac{21}{6}\)

A = \(\dfrac{175}{6}\)

Ta nhóm các thừa số có tổng là 10:

1 + 9 + 2 + 8 + 3 + 7 + 4 + 5 + 5

= ( 1 + 9 ) + ( 2 + 8 ) + ( 3 + 7 ) + ( 5 + 5 ) + 4

= 10 + 10 + 10 + 10 + 4

= 10 x 4 + 4

= 40 + 4

= 44

1+9+2+8+3+7+4+5+5

=(1+9)+(2+8)+(3+7)+(5+5)+4

=10+10+10+10+4

=44

1 + 9 + 2 + 8 + 3 + 7 + 4 + 5 + 5

= ( 1 + 9 ) + ( 2 + 8 ) + ( 3 + 7 ) + ( 4 + 6 ) + ( 5 + 5 )

= 10 + 10 + 10 + 10 + 10 

= 50 

\(\frac{999}{1000}+\frac{998}{1000}+......+\frac{1}{1000}\)

\(=\frac{999+998+997+........+1}{1000}\)

\(=\frac{499500}{1000}=\frac{999}{2}\)

\(\frac{999}{1000}+\frac{998}{1000}+.....+\frac{1}{1000}\)

= \(\frac{999+998+.....+1}{1000}\)(cách tính phép tính này rất đơn giản,chỉ việc lấy(999 + 1) x 999 : 2 = ?)

= \(\frac{499500}{1000}=\frac{999}{2}\)

k mình đi! mình trả lời rõ ràng nhất mà

1/1000 + ... + 997/1000 + 998/1000 + 999/1000 = ( 1 + ... + 997 + 998 + 999 ) / 1000 = 499500/1000 = 4995/10

theo thứ tự 1,6/4=1 và 1/2,2,5/2 ,500

các bạn giải ra giup mik đi

??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
??????????????????|


 

\(\frac{2}{3}+\frac{1}{3}=1=\frac{2}{2}\)

\(\frac{3}{4}+\frac{2}{4}+\frac{1}{4}=\frac{6}{4}=\frac{3}{2}\);

\(\frac{4}{5}+\frac{3}{5}+\frac{2}{5}+\frac{1}{5}=2=\frac{4}{2}\)

;\(\frac{5}{6}+\frac{4}{6}+\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{15}{6}=\frac{5}{2}\)

Tổng quát:

\(\frac{n-1}{n}+\frac{n-2}{n}+...+\frac{2}{n}+\frac{1}{n}\)(\(n\in N\)) \(=\frac{n-1}{2}\)

Áp dụng:

\(\frac{999}{1000}+\frac{998}{1000}+\frac{997}{1000}+...+\frac{1}{1000}=\frac{999}{2}\).

Xem bài mình đúng không?

đây là bài tập cô tớ giao

tổng nào

2010¹⁰+2

2012¹²-2

a: \(6\sqrt{3}=\sqrt{108}>\sqrt{54}=3\sqrt{6}\)

\(\Rightarrow5^{6\sqrt{3}}>5^{3\sqrt{6}}\)

b: \(\sqrt{2}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}+\dfrac{2}{3}}=2^{\dfrac{7}{6}}\)

\(\left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}=2^{\left(-1\right)\cdot\left(-\dfrac{4}{3}\right)}=2^{\dfrac{4}{3}}\)

mà \(\dfrac{7}{6}< \dfrac{8}{6}=\dfrac{4}{3}\).

nên \(\sqrt{2}\cdot2^{\dfrac{2}{3}}< \left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}\).