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Bài 2: Mỗi xe ô tô có 4 bánh xe . Hỏi 5 xe ô tô như thế có bao nhiêu bánh xe ?
Bài giải
5 xe ô tô như thế có số bánh xe là :
4 x 5= 20 (bánh xe )
Đáp số : 20 bánh xe
Giải:
A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
A=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20
A=1/2-1/20
A=9/20
D=1/3+1/6+1/12+1/24+1/48
D=1/3+1/2.3+1/3.4+1/4.6+1/6.8
D=1/3+1/2-1/3+1/3-1/4+1/2.(2/4.6+2/6.8)
D=1/3+1/2-1/4+1/2.(1/4-1/6+1/6-1/8)
D=1/3+1/4+1/2.(1/4-1/8)
D=1/3+1/4+1/2.1/8
D=1/3+1/4+1/16
D=31/48
F=0,5-1/3-0,4-4/7-1/6+4/35-1/41
F=1/2-1/3-2/5-4/7-1/6+4/35-1/41
F=1/6-(-6/35)-1/6+4/35-1/41
F=(1/6-1/6)+(6/35+4/35)-1/41
F=0+2/7-1/41
F=2/7+1/41
F=75/287
Chúc bạn học tốt!
\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)
\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)
a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)
=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )
=> \(14.\frac{27}{28}=\frac{419}{28}\)
b, \(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)
=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)
bài 2 :
( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632
=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632
=> 211x + 22366 = 23632
=> 211x = 23632 - 22366
=> 211x = 1266
=> x = 1266 : 211
x = 6
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)
ta có
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)\(-4\)
\(\Rightarrow\left(\frac{x+1}{212}+1\right)+\left(\frac{x+2}{211}+1\right)+\left(\frac{x+3}{210}+1\right)+\left(\frac{x+4}{209}+1\right)=-4+4\)
=> \(\frac{x+1+212}{212}+\frac{x+2+211}{211}+\frac{x+3+210}{210}+\frac{x+4+209}{209}\) =\(0\)
=> \(\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}\)=\(0\)
=> (x+213) \(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)=0
mà\(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)\(\ne0\)
=>x+213=0 => x=-213
vậy x= -213
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)
\(\Rightarrow\frac{x+1}{212}+1+\frac{x+2}{211}+1+\frac{x+3}{210}+1+\frac{x+4}{209}+1=-4+4=0\)
\(\Rightarrow\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}=0\)
\(\Rightarrow\left(x+213\right)\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)=0\)
\(\Rightarrow x+213=0\Leftrightarrow x=-213\)
\(\frac{2319}{2010}-\frac{1789}{2000}-\frac{309}{2010}+\frac{2009}{2010}-\frac{211}{2000}\)
\(=\left(\frac{2319}{2010}-\frac{309}{2010}+\frac{2009}{2010}\right)-\left(\frac{1789}{2000}-\frac{211}{2000}\right)\)
\(=\frac{4019}{2010}-\frac{789}{1000}\)
=))
Hình như câu này tớ đã gặp đâu đó trong đề thi HSG rồi!
\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)
\(=\frac{1}{2}\div4=\frac{1}{8}\)