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Ta có: \(\frac{1}{50}\) >\(\frac{1}{100}\)
\(\frac{1}{51}\)>\(\frac{1}{100}\)
\(\frac{1}{52}\)>\(\frac{1}{100}\)
..................
\(\frac{1}{99}\)>\(\frac{1}{100}\)
=>\(\frac{1}{50}\)+\(\frac{1}{51}\)+.............+\(\frac{1}{99}\)>\(\frac{1}{100}\).50=\(\frac{1}{2}\)(50 là số số hạng của S nha)
=>S>\(\frac{1}{2}\)
\(A=47.36+64.47+15\)
\(A=47.\left(36+64\right)+15\)
\(A=47.100+15\)
\(A=4700+15\)
\(A=4715\)
\(B=27+35+65+73+75\)
\(B=\left(27+73\right)+\left(35+65\right)+75\)
\(B=100+100+75\)
\(B=275\)
\(C=37+37.15+84.37\)
\(C=37.\left(1+15+84\right)\)
\(C=37.100\)
\(C=3700\)
\(D=\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+\frac{1}{23.24}\)
\(D=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+\frac{1}{23}-\frac{1}{24}\)
\(D=\frac{1}{20}-\frac{1}{24}\)
\(D=\frac{24}{480}-\frac{20}{480}\)
\(D=\frac{4}{480}=\frac{1}{120}\)
\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(E=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(E=1-\frac{1}{50}\)
\(E=\frac{49}{50}\)
a: Ta có
A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)
⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng
⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)
⇒ A > 1
vậy A > 1
b: ta có
S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))
⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))
⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)
⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)
⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)
vậy S > \(\dfrac{1}{2}\)
trả lời thế này chắc được điểm cao đó :
Ta thấy : \(\frac{5}{20}>\frac{5}{24}\); \(\frac{5}{21}>\frac{5}{24}\); \(\frac{5}{22}>\frac{5}{24}\); \(\frac{5}{23}>\frac{5}{24}\); \(\frac{5}{24}=\frac{5}{24}\)
\(\Rightarrow\)\(S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>\frac{5}{24}+\frac{5}{24}+\frac{5}{24}+\frac{5}{24}+\frac{5}{24}=\frac{5}{24}.5=\frac{25}{24}\)
\(S>\frac{25}{24}>\frac{24}{24}=1\)
\(\Rightarrow S>1\)
Ta có :
1<5/24x5
Mà 5/20>5/24
5/21>5/24
5/22>5/24
5/23>5/24
5/24=5/24
=>5/20+5/21+5/22+5/23+5/24>5x5/24
S>1
A = 47 x 36 + 64 x 47 + 15
A= 47 x ( 64 + 36 ) + 15 = 47 x 100 + 15 = 4700 + 15 = 4715
vậy A= 4715
B= 27+35 + 65 + 73+ 75
B= (27+ 73) + ( 35 + 65) +75
B= 100 +100 +75 = 275
vậy B= 275
C= 37 +37 x 15 +37 x 84
C= 37 x ( 1+15 +84 )= 37 x 100 = 3700
vậy C= 3700
D = 1/20x21 + 1/21x22 + 1/22x23 + 1/23x24
D= 1/20 - 1/21 + 1/21 - 1/22 + 1/22 - 1/23 + 1/23 - 1/24
D= 1/20 -1/24 = 1/120 vậy D= 1/120
E= 1/1x2 + 1/2x3 + ...... + 1/49x50
E= 1/1 - 1/2 + 1/2 - 1/3 +...... + 1/49 - 1/50
E = 1 - 1/50 = 49/50
vậy E= 49/50
CHÚC HOK TOT
ta thấy : 1/21>1/33;...1/30>1/33
Vậy 1/21+..+1/30>1/33+...+1/33(10 lần 1/33)
1/3=11/33
mà 1/33+..+1/33(10 lần 1/33) =10/33
Suy ra S>1/33+..+1/33(10 lần 1/33)>1/3
Vậy S>1/3
nhớ k nha bạn
viết lôn nha câu đầu la .. 1/30.>1/33