a: =100x26=2600

b: =78x10=780

c: =67x100=6700

d: =1000x575=575000

\(4\times26\times25=\left(4\times25\right)\times26=100\times26=2600\\ 63\times78-53\times78=\left(63-53\right)\times78=10\times78=780\\ 67\times92+67\times8=67\times\left(92+8\right)=67\times100=6700\\ 8\times23\times25\times125=\left(8\times125\right)\times\left(23\times25\right)=1000\times575=57500\)

a: =100x26=2600

b: =78x10=780

c: =67x100=6700

d: =1000x575=575000

=1/3(3/2*5+3/5*8+...+3/95*98)

=1/3(1/2-1/5+1/5-1/8+...+1/95-1/98)

=1/3*96/196

=32/196

=8/49

Ta có:\(A=\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{11}+...+\dfrac{31}{92}-\dfrac{32}{95}+\dfrac{32}{95}-\dfrac{33}{98}\)

                \(=\dfrac{1}{2}+\dfrac{33}{98}=\dfrac{82}{98}=\dfrac{41}{49}\)

Ta có: \(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(=\dfrac{8}{49}\)

\(A=\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+\frac{2}{8\cdot11}+...+\frac{2}{92\cdot95}+\frac{2}{95\cdot98}\)

\(A=\frac{2}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right]\)

\(A=\frac{2}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right]\)

\(A=\frac{2}{3}\left[\frac{1}{2}-\frac{1}{98}\right]=\frac{2}{3}\left[\frac{49}{98}-\frac{1}{98}\right]=\frac{2}{3}\cdot\frac{48}{98}=\frac{2}{3}\cdot\frac{24}{49}=\frac{2}{1}\cdot\frac{8}{49}=\frac{16}{49}\)

\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{92.95}+\frac{2}{95.98}\)

\(=\frac{2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(=\frac{2}{3}.\frac{24}{49}\)

\(=\frac{16}{49}\)

#)Giải :

\(A=2-\frac{2}{5}+\frac{2}{5}-\frac{2}{8}+\frac{2}{8}-\frac{2}{11}+...+\frac{2}{95}-\frac{2}{98}\)

\(A=2-\frac{2}{98}\)

\(A=1\frac{48}{49}=\frac{97}{49}\)

         #~Will~be~Pens~#

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)

=> 3A = \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)

=> 3A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)

=> 3A = \(\frac{1}{2}-\frac{1}{98}\)

=> 3A = \(\frac{24}{49}\)

=> A = \(\frac{8}{49}\)

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)

\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{1}{3}\cdot\frac{24}{49}=\frac{8}{49}\)

Sửa 95.98 thành 1/(95.98) nhá

Ta có

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\)

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)

\(3A=\frac{1}{2}-\frac{1}{98}\)

\(3A=\frac{49}{98}-\frac{1}{98}\)

\(3A=\frac{48}{98}=\frac{24}{49}\)

\(A=\frac{24}{49}\div3\)

\(A=\frac{8}{49}\)

Hok Tốt !!!!!!!!!!!!!!!!!!!!!!

\(\left(-67\right)+125+\left(-33\right)+75\)

\(=\left[\left(-67\right)+\left(-33\right)\right]+\left(125+75\right)\)

\(=100+200=300\)

_______

\(86.\left(-108\right)+86.9-86\)

\(=86.\left[\left(-108\right)+9-1\right]\)

\(=86.\left(-100\right)=-8600\)

_______

\(23.\left(-16\right)-23.84+300\)

\(=23.\left[\left(-16\right)-84\right]+300\)

\(=23.\left(-100\right)+300\)

\(=-2300+300\)

\(=-2000\)

______

\(235-5\left[\left(5^3-3^3\right):14\right]\)

\(=235-5\left[\left(125-27\right):14\right]\)

\(=235-5\left[98:14\right]\)

\(=235-5.7\)

\(=235-35\)

\(=200\)

_______

\(95-\left(129-74\right):5+2022^0\)

\(=95-55:5+1\)

\(=95-11+1\)

\(=84+1=85\)

\(#NqHahh\)

đặt

\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+..+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot97}\)

\(3A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot97}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{97}\\ 3A=\dfrac{95}{194}\\ A=\dfrac{95}{582}\)

a) ( -125) + 100 + 80 + 125 + 20

= [ ( -125)+125] + ( 80+20) + 100

= 0 + 100 + 100 = 200

b) 27 + 55 + (-17) + ( - 55)

= [ 27+(-17)] +[ 55+(-55)]

= 10 + 0 = 10

c) (-92) + (-251) + (-8) + 251

= [ (-92) + (-8)] + [ (-251)+251]

= (-100) + 0 = -100.

d) (-31) + (-95) +131 + (-5)

= [ (-31)+131] +[ (-95)+(-5)] = 100 + (-100) = 0

e) (187 - 23) - (20-180)

= 187 - 23 - 20 + 180 = (187-180) - (23-20) = 7 - 3 = 4

g) (-50+19+143) - (-79+25+48)

= -50+19+143 + 79 - 25 - 48

= [ -50-25] + (19 + 79) + 143 - 48

= -75 + 98 + 143 - 48 = 143 - 75 + ( 98 -48)

= 68 + 50 = 118