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Ta có:
A = 100 + 98 + 96 + .........+ 2 - 97 - 95 - 93 -...- 1
<=> 100 + 98 + 96 + ... + 2 - (97+ 95 + 93 + ... + 1)
<=> 50 x (100 + 2) : 2 - 33 x (97+1) : 2
<=> 2550 - 1617
<=> 933
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
a: \(\left|x\right|-\dfrac{38}{7}\left|x\right|-\dfrac{3}{4}=2\left|x\right|-\dfrac{8}{7}\)
\(\Leftrightarrow\left|x\right|\cdot\dfrac{-45}{7}=-\dfrac{8}{7}+\dfrac{3}{4}=\dfrac{-32+21}{28}=\dfrac{-11}{28}\)
=>|x|=11/180
=>x=11/180 hoặc x=-11/180
b: \(\Leftrightarrow x\cdot\dfrac{13}{4}-\dfrac{7}{6}x=\dfrac{5}{12}+\dfrac{5}{3}\)
=>25/12x=25/12
hay x=1
a) ta có:99 - 97 + 95 - 93 + 91 - 89 + ... + 7 - 5 + 3 - 1
= (99 - 97) + (95 - 93) + ... + (7 - 5) + (3 - 1)
= 25.2=50
b)(125.37.32).4
=(125.4).(37.32)
=500.1184
=100.5.1184
=100.5920
=592000
c)A=5+8+11+14+...+302
Số số hạng của A là
(302-5):3+1=100(số)
Tổng A= (302+5).100:2=15350
b)(125.37.32):4
=125.37.32:4
=125.37.8
=125.8.37
=1000.37
=37000
Sorry bạn mk nhìn nhầm dấu
\(\left(-4\right)^2.\left(-3\right)-\left[\left(-93\right)+\left(-11+8\right)^3\right]\)
\(=16.\left(-3\right)-\left[\left(-93\right)+\left(-3\right)^3\right]\)
\(=16.\left(-3\right)-\left[\left(-93\right)+-27\right]\)
\(=16.\left(-3\right)-\left(-120\right)\)
\(=-48+120=72\)
(-4)2.(-3) - [(-93) + (-11 + 8)3]
= 16.(-3) - [(-93) + (-3)3]
= (-48) - [(-93) + (-27)]
= (-48) - (-120)
= -48 + 120
= 72