1005/2002 >1009/2010 >1007/2006

k mk nha mk đang bị âm điẻm

\(\frac{1009}{2010}\) < \(\frac{1007}{2006}\) < \(\frac{1005}{2002}\)

Bạn lấy tử rồi chia cho mẫu là ra

mẫu số nào lớn nhất thi số đó lớn nhất nha b 2010 > 2006 > 2002. mình nghĩ như v

\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)

\(\Rightarrow(\frac{x-1009}{1010}-1)+\left(\frac{x-1007}{1012}-1\right)=\left(\frac{x-1010}{1009}-1\right)+\left(\frac{x-1012}{1007}-1\right)\)

\(\Rightarrow\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}\)

\(\Rightarrow\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)

Ta có

\(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\ne0\Rightarrow x-2019=0\Rightarrow x=2019\)

\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)

\(\frac{x-1009}{1010}-1+\frac{x-1007}{1012}-1=\frac{x-1010}{1009}-1+\frac{x-1012}{1007}\)\(\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}=0\)

\(\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)

1/1010 + 1/1012 - 1/1009 - 1/1007 khác 0

=> x - 2019 =0 => x = 2019

Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)

\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)

Khi đó  \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
 

Bạn xem lời giải của mình nhé:

Giải:

Bài 2:

Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)

\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)

\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)

 \(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)

Chúc bạn học tốt!

VÌ A = 1/2010 > 1/2011 > 1/2012  (1)

     B = 1/2009 <1/1007 (2) 

TỪ (1) VÀ (2)  => 1/2010 < 1/1007

VẬY A < B

Sao bạn biết

a , Ta có :     \(1-\frac{54}{59}=\frac{5}{59}\) \(=\frac{50}{590}\)    ;     \(1-\frac{541}{591}=\frac{50}{591}\)

Vì \(\frac{50}{590}>\frac{50}{591}\)nên \(\frac{54}{59}< \frac{541}{591}\)

\(A=\frac{2016^{2016}+2}{2016^{2016}-1};;B=\frac{2016^{2016}}{2016^{2016}-3}\)\(A=\frac{\left(2016^{2016}-1\right)+2+1}{2016^{2016}-1};;B=\frac{\left(2016^{2016}-3\right)+3}{2016^{2016}-3}\)\(A=1+\frac{3}{2016^{2016}-1};;B=1+\frac{3}{2016^{2016}-3}\);;Vì \(2016^{2016}-1>2016^{2016}-3\)Nên\(\frac{3}{2016^{2016}-1}< \frac{3}{2016^{2016}-3}\)Vậy \(A< B\)