Tính 1/20 +1/21+...+1/38
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\(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+.....+\frac{1}{37\cdot38}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{37}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{38}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{38}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{38}\right)\)
\(=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{38}\)
\(N=\frac{1}{20\cdot38}+\frac{1}{21\cdot37}+...+\frac{1}{38\cdot20}\)
\(\Rightarrow58N=\frac{1}{20}+\frac{1}{38}+\frac{1}{21}+\frac{1}{37}+...+\frac{1}{37}+\frac{1}{20}\)
\(=2\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{38}\right)\)
\(=2A\)
\(\Rightarrow N=\frac{2}{58}M\)
\(\Rightarrow\frac{M}{N}=29\)là số nguyên.
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
Lời giải:
a. $=-(37+63)+[25+(-25)]+(-9)=-100+0+(-9)=-(100+9)=-109$
b. $=[1+(-3)]+[5+(-7)]+....+[21+(-23)]$
$=\underbrace{(-2)+(-2)+....+(-2)}_{6}=(-2).6=-12$
c. $=-(280+20)+[-(79+21)]=-300+(-100)=-(300+100)=-400$
d. $=[-(27+43)]+[-(208+102)]=-70+(-310)=-(70+310)=-380$
e. $=(38+120)-(12+46)=158-58=100$
f. $=9+15+11+24=(9+11)+(15+24)=20+39=59$
21)
\(\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right).....\left(1+\dfrac{1}{9999}\right)\\ =\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{10000}{9999}\\ =\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{100.100}{99.101}\\ =\dfrac{2.3.4.....100}{1.2.3.....99}.\dfrac{2.3.4.....100}{3.4.5.....101}\\ =100.\dfrac{2}{101}\\ =\dfrac{200}{101}\)
\(\frac{5}{6}:7:\frac{20}{21}+\frac{5}{6}:7:\frac{20}{21}=\frac{5}{6}:7:\frac{20}{21}x2=\frac{1}{4}\)
mẫu số có ssh là
(38-20):1+1=19 số
tổng của tử số là
(38+20)x19:2=551
Ta được 1/551