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a/ Ta có :
\(10A=\frac{10\left(10^{50}+1\right)}{10^{51}+1}=\frac{10^{51}+10}{10^{51}+1}=\frac{10^{51}+1}{10^{51}+1}+\frac{9}{10^{51}+1}=1+\frac{9}{10^{51}+1}\)
\(10B=\frac{10\left(10^{51}+1\right)}{10^{52}+1}=\frac{10^{52}+10}{10^{52}+1}=\frac{10^{52}+1}{10^{52}+1}+\frac{9}{10^{52}+1}=1+\frac{9}{10^{52}+1}\)
Vì \(\frac{9}{10^{51}+1}>\frac{9}{10^{52}+1}\Leftrightarrow10A>10B\Leftrightarrow A>B\)
Vậy...
b/ Mình sửa lại một chút nhé :>
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}-3=0\)
\(\Leftrightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)=0\)
\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
Vậy...
c/ Đặt :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{1999.2000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{1999}-\frac{1}{2000}\)
\(=1-\frac{1}{2000}\)
\(=\frac{1999}{2000}\)
Vậy..
\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{49^2}-\frac{1}{50^2}\)
\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)
\(T=\frac{1}{\left(2-1\right)\left(2+1\right)}+\frac{1}{\left(3-1\right)\left(3+1\right)}+...+\frac{1}{\left(50-1\right)\left(50+1\right)}\)
\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)
Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T
JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)
Biến đổi vp của đẳng thức :
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}-2\left[\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right]\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\)
B= (1/2-1/3) + (1/3-1/4) + (1/4-1/5)+...+( 1/99-1/100)
B = (1/2-1/3) + (1/3 - 1/4) + (1/4 - 1/5)+...+ (1/99 + 1/100)
B= 1/2 +1/100=51/100
k mk nhóe
sai thì chỉ mk nhoa
a)A=1/51+1/52+...+1/100
=>A>1/100+1/100+...+1/100
=>A>50/100(vì có 50 số hạng)
=> A>1/2
b)Ta có:
B=1/2.3+1/3.4+...+1/99.100
=> B=1/2-1/3+1/3-1/4+...+1/99-1/100
=> B=1/2-1/100
Mà 1/100>0
=> B<1/2
=> B<1/2<A
=>B<A
S = \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.50=\frac{1}{2}\)
Kết luận vậy S > 1/2
S=1/2