583.931+69.(170+583)+31.170
ghi đầy đủ nhá
làm giúp ik aaaaa
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Bạn kiểm tra lại giùm mình đi nha:
15.37.4+120.21+21.5.12 16.64+76.34
=15.37.4+(6.7).4.(5.3)+(3.5).(7.3).4 =
=15.37.4+42.4.15+15.21.4
=15.4.(37+42+21)
=60.100
=6000
\(a,2\times31\times12+4\times6\times42+8\times27\times3\)
\(=24\times31+24\times42+24\times27\)
\(=24\times\left(31+42+27\right)\)
\(=24\times100\)
\(=2400\)
\(b,583.931+69.\left(170+583\right)+31.170\)
\(=583.931+69.170+69.583+31.170\)
\(=583.\left(931+69\right)+170.\left(69+31\right)\)
\(=583.1000+170.100\)
\(=583000+17000\)
\(=600000\)
\(c,732.976+24.\left(680+732\right)+76.680\)
\(=732.976+24.680+24.732+76.680\)
\(=732.\left(976+24\right)+680.\left(24+76\right)\)
\(=732.1000+680.100\)
\(=732000+68000\)
\(=800000\)
Học tốt
Đặt A=\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)
=(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\))+(\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\))+...+\(\dfrac{1}{70}\)
Nhận xét:
\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\)>\(\dfrac{1}{20}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{1}{20}\)=\(\dfrac{10}{20}\)=\(\dfrac{1}{2}\)
\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\)>\(\dfrac{1}{30}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{30}\)=\(\dfrac{10}{30}\)=\(\dfrac{1}{3}\)
\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\)>\(\dfrac{1}{60}\)+\(\dfrac{1}{60}\)+...+\(\dfrac{1}{60}\)=\(\dfrac{30}{60}\)=\(\dfrac{1}{2}\)
=>A>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)+\(\dfrac{1}{61}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)=\(\dfrac{4}{3}\)
=>A>\(\dfrac{4}{3}\)
Vậy: \(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{4}{3}\) (ĐPCM)
Thấy đúng cho 1 tick và 1 follow nha!
Chúc bạn học tốt!
dễ thấy vế trái luôn>0 nên 6x>0=> x>0
x>0, bỏ dấu trị tuyệt đối ra ta đc 4x+10=6x
x=5
chúc bạn học giỏi, ăn Tết đc ngon, hehe -_-
HYC-30/1/2022
Answer:
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=6x\)
Có \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+1+x+2+x+3+x+4=6x\)
\(\Rightarrow4x+10=6x\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
Ta có: 1 + 2 + 3 + 4 + ... + 2004 + 2005 + 2006 + 2007
Dãy đó có số số hạng là:
( 2007 - 1 ) : 2 = 1003 ( số hạng )
Tổng của các số trên là:
( 1 + 2007 ) x 1003 : 2 = 4028048
Đáp số: 4028048
\(583.931+69\left(170+583\right)+31.170\)
\(=583.931+69.170+69.583+31.170\)
\(=593\left(931+69\right)+170\left(69+31\right)\)
\(=593.100.10+170.100\)
\(=100\left(593.10+170\right)\)
\(=100.6000=600000\)