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S=(1/31+1/32+1/33+...+1/40)+(1/41+1/42+1/43+...+1/50)+(1/51+1/52+1/53+...+1/60)"10 sống hạng mỗi ngoặc"
S<1/30 x 10+1/40 x 10+1/50 x 10
S<1/3+1/4+1/5=47/60<48/60=4/5
Học tốt~
A = 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2017. 2019
= ( 1 - 1/3 ) + ( 1/3 - 1/5 ) + ... + (1/2017 - 1/2019 )
= 1 - 1/2019
= 2018/2019
S = 1/31 + 1/32 +...+ 1/60
Ta có các phân số : 1/31, 1/32, ..., 1/59 đều lớn hơn 1/60
Nên S > 1/60 + 1/60 + 1/60 +...+ 1/60 ( có tất cả 30 phân số )
= 30/60 = 1/2
Vì 1/2 < 4/5 nên S <4/5
Vậy, chứng tỏ S < 4/5
Chúc bạn học tốt !
a, \(\frac{3}{5}+\frac{-4}{15}=\frac{9}{15}-\frac{4}{15}=\frac{5}{15}=\frac{1}{3}\)
b, \(\frac{-1}{3}+\frac{2}{5}+\frac{2}{15}=\frac{-5}{15}+\frac{6}{15}+\frac{2}{15}=\frac{3}{15}=\frac{1}{5}\)
c, \(\frac{-3}{5}+\frac{7}{21}+\frac{-4}{5}+\frac{7}{5}=\frac{-3}{5}+\frac{1}{3}+\frac{-4}{5}+\frac{7}{5}=\left(\frac{-3}{5}+\frac{-4}{5}+\frac{7}{5}\right)+\frac{1}{3}=\frac{1}{3}\)
d, \(\frac{2}{7}+\frac{1}{9}+\frac{3}{7}+\frac{5}{9}+\frac{-5}{6}=\left(\frac{2}{7}+\frac{3}{7}\right)+\left(\frac{1}{9}+\frac{5}{9}\right)+\frac{-5}{6}=\frac{5}{7}+\frac{6}{9}+\frac{-5}{6}=\frac{90}{126}+\frac{84}{126}+\frac{-105}{126}=\frac{69}{126}=\frac{23}{42}\)
e, \(\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}=\left(-1\right)+1+\frac{-1}{5}=\frac{-1}{5}\)
f, \(\frac{-3}{31}+\frac{-6}{17}+\frac{1}{25}+\frac{-28}{31}+\frac{-1}{17}+\frac{-1}{5}=\left(\frac{-3}{31}+\frac{-28}{31}\right)+\left(\frac{-6}{17}+\frac{-1}{17}\right)+\left(\frac{1}{25}+\frac{-1}{5}\right)=\left(-1\right)+\frac{-7}{17}+\frac{-4}{25}=\frac{-425}{425}+\frac{-175}{425}+\frac{-68}{425}=\frac{-668}{425}\)
Chúc bn học tốt
\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
Ta có:
M=\(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\)
M=\(\frac{1.3....99}{2.4....100}\)
Lại có:
N=\(\frac{2}{3}.\frac{4}{5}....\frac{100}{101}\)
N=\(\frac{2.4....100}{3.5....101}\)
\(\Rightarrow\)M.N=\(\frac{1.2.3......99.100}{2.3.4......100.101}\)
\(\Rightarrow\)M.N=\(\frac{1}{101}\)
Ta có:
\(A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)
\(A>\dfrac{1}{40}.10+\dfrac{1}{50}.10+\dfrac{1}{60}.10=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{3}{5}\)
Vậy \(A>\dfrac{3}{5}\)
Ta có:
\(A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)\(A< \dfrac{1}{31}.10+\dfrac{1}{41}.10+\dfrac{1}{51}.10< \dfrac{4}{5}\)
Vậy \(A< \dfrac{4}{5}\)
Do đó: \(\dfrac{3}{5}< A< \dfrac{4}{5}\)