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\(\frac{50}{111}>\frac{1}{4};\frac{50}{112}>\frac{1}{4};\frac{50}{113}>\frac{1}{4};\frac{50}{114}>\frac{1}{4}\)
\(A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}>\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)(1)
\(\frac{50}{111}< \frac{1}{2};\frac{50}{112}< \frac{1}{2};\frac{50}{113}< \frac{1}{2};\frac{50}{114}< \frac{1}{2}\)
\(\Rightarrow A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}< \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)(2)
từ (1) và (2) \(\Rightarrow1< A< 2\)
50/111 < 50/100
50/112 < 50/100
50/113 < 50/100
50/114 < 50/100
=> A < 200/100 => A < 2
50/111 > 50/200
50/112 > 50/200
50/113 > 50/200
50/114 > 50/200
=> A > 200/200 => A > 1
Vậy 1 < A < 2
AI THẤY OK ỦNG HỘ NHÉ
Ta có :
\(A=\dfrac{50}{111}+\dfrac{50}{112}+\dfrac{50}{113}+\dfrac{50}{114}\)
Ta thấy :
\(\dfrac{50}{111}>\dfrac{50}{200}\)
\(\dfrac{50}{112}>\dfrac{50}{200}\)
\(\dfrac{50}{113}>\dfrac{50}{200}\)
\(\dfrac{50}{114}>\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}.4=1\) \(\left(1\right)\)
Mặt khác :
\(\dfrac{50}{111}< \dfrac{50}{100}\)
\(\dfrac{50}{112}< \dfrac{50}{100}\)
\(\dfrac{50}{113}< \dfrac{50}{100}\)
\(\dfrac{50}{114}< \dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}.4=2\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow1< A< 2\rightarrowđpcm\)
A<50/100+50/100+50/100+50/100=4.50/100=2
=>A<2
A>4.50/150=4/3+1+1/3>1
=>dccm
A = 1/12 + 1/22 + 1/32 + ... + 1/502
A = 1/1.1 + 1/2.2 + 1/3.3 + ... + 1/50.50
A < 1/1 + 1/1.2 + 1/2.3 + ... + 1/49.50
A < 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50
A < 2 - 1/50 < 2
Chứng tỏ A < 2
Đặt B=1/1+1/1.2+...+1/49.50
Ta có:
A=1/1^2+1/2^2+...+1/50^2<B=1/1+1/1.2+...+1/49.50 (1)
Mà B=1/1+1/1.2+...+1/49.50
=1+1-1/2+1/2-1/3+...+1/49-1/50
=2-1/50 <2 (2)
Từ (1) và (2) =>A<B<2
=>A<2
Ta có:
\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}\)
\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}\)
Vì \(10^{50}-1>10^{50}-3\Rightarrow\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)(2 phân số có cùng tử số, mẫu số của phân số nào lớn hơn thì phân
số đó nhỏ hơn)
\(\Rightarrow1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\Rightarrow A< B\)
\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}.\)
\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}.\)
Do 1050-1 > 1050-3 ; => \(1+\frac{3}{10^{50}-3}>1+\frac{3}{10^{50}-1}\)
=> B > A
Ta có :
\(\frac{50}{111}>\frac{50}{200}\)
\(\frac{50}{112}>\frac{50}{200}\)
\(\frac{50}{113}>\frac{50}{200}\)
\(\frac{50}{114}>\frac{50}{200}\)
\(\Rightarrow A>\frac{50}{200}+\frac{50}{200}+\frac{50}{200}+\frac{50}{200}\)hay \(A>\frac{50}{200}.4\left(1\right)\)
Mặt khác :
\(\frac{50}{111}< \frac{50}{100}\)
\(\frac{50}{112}< \frac{50}{100}\)
\(\frac{50}{113}< \frac{50}{100}\)
\(\frac{50}{114}< \frac{50}{100}\)
\(\Rightarrow A< \frac{50}{100}+\frac{50}{100}+\frac{50}{100}+\frac{50}{100}\)hay \(A< \frac{50}{100}.4\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\Rightarrow1< A< 2\left(đpcm\right)\)