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(-2) . 29 + (-2) . (-99) + (-2) . (-30)
= (-2) . [29 + (-99) + (-30)]
= (-2) . [ (-70)+(-30)]………………
= (-2).(-100)=200………………….
(-2) . 29 + (-2) . (-99) + (-2) . (-30)
= (-2) . [ 29 + (-99) + (-30)]
=(-2) . (-100)
=200
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
b, ( cái (99) là -99 hay +99)
\(c,=\left[\left(-25\right).\left(-4\right)\right].\left[9.\left(-7\right)\right]\\ =100.\left(-63\right)=-6300\\ d,=37.\left(-84\right)+37.\left(-16\right)\\ =37.\left[\left(-84\right)+\left(-16\right)\right]\\ =37.\left(-100\right)=-3700\)
b) (-2).29+(-2).(99)+(-2).(-30)
=1.(-2).29+(-2).(99)+(-2)
=-2. (1+99+30)
=-2.130
=-260
c)(-25).9.(-4).(-7)
=[-25.(-4)].[9.(-7)]
=100.(-63)
=-6300
d) (-37).84+37.(-16)
=(-37+37).[84+(-16)]
=0.[84+(-16)]
=0
\(\left( { - 2} \right).29 + \left( { - 2} \right).\left( { - 99} \right)\)\( + \left( { - 2} \right).\left( { - 30} \right)\)\( = \left( { - 2} \right)\left( {29 - 99 - 30} \right)\)\( = \left( { - 2} \right).\left( { - 100} \right) = 200\)
\(a)A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)
\(A=\dfrac{2.3}{16}\)
\(=\dfrac{3}{8}\)
\(b)B=\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}\)
\(B=\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{56}\right)\)
\(B=6-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)
\(B=6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(B=6-\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)
\(B=\dfrac{48}{8}-\dfrac{3}{8}\)
\(B=\dfrac{45}{8}\)
(-2).99 + (-2).(-99) + (-2).(-30)
= 99.(-2 + 2) + 60
= 99.0 + 60
= 0 + 60
= 60
A = 1+ 2 + 3 + 4 + 5 + ... + 99
Số các số hạng của A là : ( 99 -1 ) : 1 + 1 = 99 ( số hạng )
A = ( 1+ 99 ) . 99 : 2 = 4950
Vậy A = 4950
B = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{99}\)
B = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.11}\)
????????????????????????????????? Mình nghĩ đầu bài phải là : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
A = 1 + 2 + 3 + 4 + 5 + ... + 99
Số số hạng của A là:
(99 - 1) : 1 + 1 = 99 (số hạng)
Tổng dãy số trên là:
(99 + 1) x 100 : 2 = 5000 (số hạng)
phần B có vấn đề nha :)
\(\frac{1}{2}+\frac{5}{6}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)\)
\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=6-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=6-\left(1-\frac{1}{7}\right)=6-\frac{6}{7}=\frac{36}{7}\)
a) 19 + (29 - 9*37) - (63*9 - 29*99)
= 19 + 29 - 9*37 - 63*9 + 29*99
= 19 + 29(1 + 99) - 9(37 + 63)
= 19 + 29*100 - 9*100
= 19 + 100(29 - 9)
= 19 + 100*20
= 19 + 2000 = 2019
b) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
= \(\frac{2^6+2^5+2^4+2^3+2^2+2+1}{2^7}\)
= \(\frac{64+32+16+8+4+2+1}{128}\) = \(\frac{127}{128}\)
(-20).29+(-2).(-99)+(-2).(-30)
=(-2).10.29+(-2).(-99)+(-2).(-30)
=(-2).(290-99-30)
=(-2).161
=-322
HT