mỗi  số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15

ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

hay tong tren be hon 15

bạn ơi bạn ngu vcl lun là>

Đề đúng phải là:

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)

Cộng mỗi phân thức thêm 1, quy đồng rồi chuyển sang 1 vế ta được:

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+2015}{2003}=0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Mà BT tích sau luôn nhỏ hơn 0

=> x+2015=0 => x = -2015

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)( như này đúng không ? :)) )

<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+10}{2005}+1\right)+\left(\frac{x+11}{2004}+1\right)+\left(\frac{x+12}{2003}+1\right)\)

<=> \(\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}+\frac{x+3+2012}{2012}=\frac{x+10+2005}{2005}+\frac{x+11+2004}{2004}+\frac{x+12+2003}{2003}\)

<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}=\frac{x+2015}{2005}+\frac{x+2015}{2004}+\frac{x+12}{2003}\)

<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+12}{2003}=0\)

<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)

=> x + 2015 = 0

=> x = -2015

\(B=\frac{2003+2004}{2004+2005}=\frac{2003}{2004+2005}+\frac{2004}{2004+2005}\)

Ta có: \(\frac{2003}{2004}>\frac{2003}{2004+2005}\)

          \(\frac{2004}{2005}>\frac{2004}{2004+2005}\)

\(\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003+2004}{2004+2005}\)

\(A>B\)

Vậy A>B

đáp án là A>B

chúc bn hok tốt!

                                 \(\text{ Bài giải}\)

\(A=\frac{2003}{2004}+\frac{2004}{2005}=0,999500998 + 0,999501247=1.99900225\)

\(B=\frac{2003+2004}{2004+2005}=\frac{4007}{4009}=0,999501122\)

\(\text{Vì : }1,99900224>0,999501122\text{ nên }A>B\)

        \(\text{Vậy : }A>B\)

a)\(\frac{2013}{2015}< \frac{2014}{2016}\)

b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)

ta có tính chất \(\frac{a}{b}\)>1 suy ra \(\frac{a.m}{b.m}\).........

Ta có:

\(a=\frac{2013}{2014}=1-\frac{1}{2014}\)

\(b=\frac{2003}{2004}=1-\frac{1}{2004}\)

Vì \(\frac{1}{2014}< \frac{1}{2004}\Rightarrow1-\frac{1}{2014}>1-\frac{1}{2004}\)

Vậy \(a>b\)

Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015}\) (1)
\(\frac{2014}{2015}>\frac{2014}{2014+2015}\) (2)
ộng caác bất đẳng thứa (1) và (2) vào vế với vế:
\(\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\Rightarrow A>B\)

cảm ơn các bạn                         

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Ta có:

\(\frac{20032003}{20022002}=\frac{20032003:10001}{20022002:10001}=\frac{2003}{2002}\)

\(\frac{2014}{2013}=\frac{2014.182}{366366}=\frac{366548}{366366}\)

\(\frac{2003}{2002}=\frac{2003.182}{366366}=\frac{364546}{366366}\)

\(\Rightarrow\frac{2014}{2013}>\frac{2003}{2002}\)

ps 2014/2013 lon hon

hay ờ nha