Ta có :

3 + ( - 12 ) + 13 + ( -22 ) + ..........+ 93 + ( - 102 )

= [ 3 + ( -12 ) ] + [ 13 + ( -22 ) ] + ... + [ 93 + ( -102 ) ]

= -9 + ( -9 ) + .... + ( -9 )

= ( -9 ) . 10

= -90

ai tra loi nhanh va dung minh se k cho

\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\\ =\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\\ =6+6+6+...+6\left(10\text{ số }6\right)\\ =6\cdot10\\ =60\)

\(b,3+\left(-12\right)+13+\left(-22\right)+23+\left(-32\right)+...+93+\left(-102\right)\\ =\left[3+\left(-12\right)\right]+\left[13+\left(-22\right)\right]+\left[23+\left(-32\right)\right]+...+\left[93+\left(-102\right)\right]\\ =\left(-9\right)+\left(-9\right)+\left(-9\right)+...+\left(-9\right)\left(10\text{ số }-9\right)\\ =\left(-9\right)\cdot10\\ =-90\)

\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\)

\(=\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\)

\(=6+6+6+...+6\) (10 số 6)

\(=6.10=60\)

\(\)

\(a)-5+11+(-15)+21+(-25)+31+.......+(-95)+101\)

\(=(-5+11)+(-15+21)+(-25+31)+.....+(-95+101)\)

\(=6+6+6+..........+6\)(10 số 6)

\(=6.10\)

\(=60\)

Vậy....

Câu b tương tự

a) Số các số hạng là:

(199 - 11) + 1 = 189 số

Tổng là:

189 x (199 + 11) : 2 = 19845

b) Số các số hạng là:

(99 - 1):2 + 1 = 50 số

Tổng là:

50 x (99 + 1):2 =2500   

a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)

= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)

= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)

= 1 - 1 - \(\dfrac{11}{25}\)

= - \(\dfrac{11}{25}\)

b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)

= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))

= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)

= - \(\dfrac{499}{35}\)

c, \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{48}\)+ ...+ \(\dfrac{1}{2400}\)

= \(\dfrac{1}{2\times4}\)  +\(\dfrac{1}{4\times6}\) + \(\dfrac{1}{6\times8}\) +...+ \(\dfrac{1}{48\times50}\)

= \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{2\times4}\) +\(\dfrac{2}{4\times6}\)+ \(\dfrac{2}{6\times8}\) +...+ \(\dfrac{2}{48\times50}\))

= \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{48}\) - \(\dfrac{1}{50}\))

= \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{12}{25}\)

= \(\dfrac{6}{25}\)

a) \(100+98+96+...+2-97-95-93-...-3\)

= \(100+98+\left(96-97\right)+\left(94-95\right)+...+\left(2-3\right)\)

= \(100+98-95\) = \(103\)

b) \(2-4-6+8+10-12-14+16+...-102+104\)

= \(\left(2-4\right)+\left(-6+8\right)+\left(10-12\right)+\left(-14+16\right)+...+\left(-102+104\right)\)

= \(-2+2-2+2-2+...+2\) = \(0\)

c) \(1+2-3-4+5+6-7-8+9+10-11-12+...-111-112+113+114\)

= \(\left(1+2\right)-\left(3+4\right)+\left(5+6\right)-\left(7+8\right)+...\left(113+114\right)\)

= \(3-7+11-15+19-23+...+219-223+227\)

= \(\left(3-7\right)+\left(11-15\right)+\left(19-23\right)+...+\left(219-223\right)+227\)

= \(-4-4-4-4-...-4+227\)

= \(54\left(-4\right)+227\) = \(-216+227\) = \(11\)