1/2 + 1/3 + 1/4 + 1/5 + 4/5 + 3/4 + 2/3 + 1/2

= (1/2 + 1/2) + (1/3 + 2/3) + (1/4 + 3/4) + (1/5 + 4/5)

= 1 + 1 + 1 + 1

= 4

=(1-2)+(3-4)+...+(99-100)+101

=(-50)+101

=51

$1-2+3-4+5-6+.....+97-98+99-100+101$

$= (1-2) + (3-4) + (5-6) + ...... + (99-100) + 101

$= (-50) + 101$

$= 51$

\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{101-99}{99x101}=\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(5\dfrac{3}{4}:3+2\dfrac{1}{4}\times\dfrac{1}{3}-\dfrac{1}{8}=\left(5\dfrac{3}{4}+2\dfrac{1}{4}\right)\times\dfrac{1}{3}-\dfrac{1}{8}\\ =8\times\dfrac{1}{3}-\dfrac{1}{8}=\dfrac{8}{3}-\dfrac{1}{8}=\dfrac{61}{24}\\ ----\\ \dfrac{3}{5}:\dfrac{5}{6}:\dfrac{6}{7}:\dfrac{7}{8}+\dfrac{2}{5}+\dfrac{23}{35}\\ =\dfrac{3}{5}\times\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}+\dfrac{2}{5}+\dfrac{23}{35}=\dfrac{18}{25}\times\dfrac{4}{3}+\dfrac{2}{5}+\dfrac{23}{35}=\dfrac{353}{175}\)

1.a,=(54+45+1).113

=100.113

=11300

b,=(3/7+8/14)+(4/9+10/18)

=1+1

=2

2.a,=13/10+1/3

=49/30

b,=12/9.(1/12+1/6)

=12/9.1/4

=1/3

c,=3/4.3/2

=9/8

d,=3/2-1/3

=7/6

1:tính bằng cách thuận tiện nhất:

a)54 x 113 + 45 x 113 + 113

= 54 x 113 + 45 x 113 + 113x1

=113 x(54+45+1)

= 113x100

=1300

 b)3/7 + 4/9 + 8/14 + 10/18

=(3/7+8/14)+(4/9+10/18)

=    1           + 1

=2

=1 + (1/2×5/2×3/2)+2/3

=1 + 15/8 + 2/3

=1 + 61/24

=85/24

dặt bt trên là A ta có

A=1+(2+101)-(3+100)+..=1+0-0+..=1

Tham khảo nhé bạn :))

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