Question

[Trần Trọng Nghĩa]

Giải hộ mình bất phương trình này.

\(\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}>\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}\)

Answer 1

[Eren]

BPT \(\Leftrightarrow\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}>0\)

\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)>0\)

\(\Leftrightarrow\dfrac{x-15}{2002}+\dfrac{x-15}{2003}-\dfrac{x-15}{2004}-\dfrac{x-15}{2005}>0\)

\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)

Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}>0\)

\(\Rightarrow x-15>0\)

\(\Leftrightarrow x>15\)

Vậy bpt có nghiệm x > 15

Answer 2

[Terry Kai]

\(\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-2>\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}-2\)

\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)\)

\(-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)\)

quy đồng lên ta được:

\(\left(\dfrac{x+1987-2002}{2002}\right)+\left(\dfrac{x-1998-2003}{2003}\right)\)

\(-\left(\dfrac{x+1989-2004}{2004}\right)-\left(\dfrac{x+1990-2005}{2005}\right)>0\)

\(\Leftrightarrow\left(\dfrac{x-15}{2002}\right)+\left(\dfrac{x-15}{2003}\right)-\left(\dfrac{x-15}{2004}\right)-\left(\dfrac{x-15}{2005}\right)>0\)

đặt nhân tử chung ta được:

\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)

Vì:

\(\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\in Z\right)\) nên ta xét \(x-15>0\Rightarrow x>15\)