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\(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\)
\(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>\dfrac{1}{90}+\dfrac{1}{90}+...+\dfrac{1}{90}=\dfrac{30}{90}=\dfrac{1}{3}\)
Do đó: \(B+C>\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(đpcm)
Ta có: A= \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{90}\)
\(A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}\right)\)
A= B+C
Ta có: \(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\)
\(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>30.\dfrac{1}{60}=\dfrac{1}{2}\) (1)
Lại có: \(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>\dfrac{1}{90}+\dfrac{1}{90}+...+\dfrac{1}{90}\)
\(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>30.\dfrac{1}{90}=\dfrac{1}{3}\) (2)
Từ (1) và (2) => \(A>\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
Vậy \(A>\dfrac{5}{6}\)
Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)
Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
\(\dfrac{1}{33}>\dfrac{1}{40}\)
\(\dfrac{1}{34}>\dfrac{1}{40}\)
\(\dfrac{1}{35}>\dfrac{1}{40}\)
\(\dfrac{1}{36}>\dfrac{1}{40}\)
\(\dfrac{1}{37}>\dfrac{1}{40}\)
\(\dfrac{1}{38}>\dfrac{1}{40}\)
\(\dfrac{1}{39}>\dfrac{1}{40}\)
\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)
Vậy \(S>\dfrac{1}{4}\)
Vì \(\frac{1}{32}+\frac{1}{33}+.....+\frac{1}{89}+\frac{1}{90}>0\)
Mà\(1>\frac{2}{3}\)
=>\(1+\frac{1}{31}+\frac{1}{32}+.....+\frac{1}{90}>0+\frac{2}{3}\)
=>\(1+\frac{1}{31}+\frac{1}{32}+....+\frac{1}{90}>\frac{2}{3}\)
Vậy......
(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+ 1/53 +1/54+ 1/55 +1/56+ 1/57 +1/58 + 1/59 +1/60)
E = 1/31+1/32+...+1/60
E > 1/40+1/40+...+1/40+1/41+1/42+...+1/60
E > 20/40+1/41+1/42+...+1/60
E > 1/2+1/60+1/60+...+1/60
E > 1/2 + 1/3 = 5/6
Mà 5/6 > 4/5
=> E > 4/5
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{59.60}\)
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}\)
\(B=\left(1+\frac{1}{3}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{59}+\frac{1}{60}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{30}\right)\)
\(B=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}=A\)
chia từng khoảng rồi so sánh em ạ