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13 tháng 10 2019

Ta có:

1 + 3 có 2 số hạng   => 1 + 3 = 2^2

1 + 3 + 5 có ( 5 - 1 ) : 2 +1 = 3 số hạng =>  1 + 3 + 5 = (5 + 1 ). 3 : 2 = 3^2

1 + 3 + 5 + 7 có: ( 7 - 1 ) : 2 + 1 =4 số hạng => 1 + 3 + 5 + 7 = ( 7 + 1 ) .4 : 2 = 4^2

...

1 + 3 + 5 + 7 +... + 101 có ( 101 -1 ) : 2 + 1 =51 số hạng => 1 + 3 + 5 + 7 +... + 101 = ( 101 + 1 ) . 51 : 2 =51^2

=> \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{51^2}\)

\(< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\)

\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{51}\right)< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

=> B < 3/4

       

 

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-(x+3/5)+4/5

=1/7-2x-3/5+4/5

=-2x+12/35

b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

-3/5<x<1/7

nên x-1/7<0; x+3/5>0

\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0

\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-(x+3/5)+4/5

=1/7-2x-3/5+4/5

=-2x+12/35

b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

-3/5<x<1/7

nên x-1/7<0; x+3/5>0

\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0

\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)

ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều 

25 tháng 7 2019

Bài 2:

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

= \(\frac{2}{3}\) - 5 = \(\frac{-13}{3}\)

b) (\(\frac{-1}{3}\) + \(\frac{5}{6}\)) . 11 - 7 = \(\frac{1}{2}\). 11 - 7

= \(\frac{11}{2}\)- 7 = \(\frac{-3}{2}\)

c) \(\frac{-5}{9}\). \(\frac{3}{11}\)+ (\(\frac{-13}{18}\)) . \(\frac{3}{11}\)= \(\frac{3}{11}\). (\(\frac{-5}{9}\)- \(\frac{13}{18}\))

= \(\frac{3}{11}\). (\(\frac{-23}{18}\))= \(\frac{-23}{66}\)

d) \(\frac{-2}{3}\) . \(\frac{3}{11}\)+ (\(\frac{-16}{9}\)) . \(\frac{3}{11}\)= \(\frac{3}{11}\). (\(\frac{-2}{3}\)- \(\frac{16}{9}\))

= \(\frac{3}{11}\). (\(\frac{-22}{9}\)) = \(\frac{-2}{3}\)

Bài 1:

a) =\(\frac{-1}{2}\)+ \(\frac{3}{5}\)- \(\frac{1}{9}\)+ \(\frac{1}{71}\) + \(\frac{2}{7}\) + \(\frac{4}{35}\)- \(\frac{7}{18}\)

= (\(\frac{-1}{2}\)- \(\frac{1}{9}\)- \(\frac{7}{18}\)) + ( \(\frac{3}{5}\)+ \(\frac{2}{7}\)+ \(\frac{4}{35}\)) + \(\frac{1}{71}\)

= -1 + 1 + \(\frac{1}{71}\)= \(\frac{1}{71}\)

b) = 3 - \(\frac{1}{4}\)+ \(\frac{2}{3}\)- 5 + \(\frac{1}{3}\)+ \(\frac{6}{5}\)- 6 + \(\frac{7}{4}\)- \(\frac{3}{2}\)

= (3 - 5 - 6) + (\(\frac{-1}{4}\)+ \(\frac{7}{4}\)- \(\frac{3}{2}\)) + (\(\frac{2}{3}\)+ \(\frac{1}{3}\)) + \(\frac{6}{5}\)

= -8 + 0 + 1 + \(\frac{6}{5}\)

= \(\frac{-29}{5}\)

28 tháng 7 2018

a, S= 1/1*2 + 1/2*3 + 1/3*4 +...+1/99*100
    S= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/99 - 1/100
    S= 1/1 - 1/100
    S= 100/100 - 1/100
    S= 99/100

b, S= 1/1*3 + 1/3*5 + 1/5*7 +...+1/99*101
    S= 1/2* (2/1*3 + 2/3*5 + 2/5*7 +...+ 2/99*101)
    S= 1/2* (1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +...+ 1/99 - 1/101)
    S= 1/2* (1/1 - 1/101)
    S= 1/2* (101/101 - 1/101)
    S= 1/2* 100/101
    S= 50/101
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