Question

Giải các phương trình sau:

a) \(\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}\)

b) \(\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}\)

c) \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

giúp mk vs

Answer 1

a) \(\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}\)

\(\Leftrightarrow\dfrac{15-x}{2000}+1+\dfrac{14-x}{2001}+1=\dfrac{13-x}{2002}+1+\dfrac{12-x}{2003}+1\)

\(\Leftrightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}=\dfrac{2015-x}{2002}+\dfrac{2015-x}{2003}\)

\(\Rightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}-\dfrac{2015-x}{2002}-\dfrac{2015-x}{2003}=0\)

\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow2015-x=0\)

<=> x=2015

Vậy phương trình có nghiệm là x=2015

Answer 2

b) \(\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}\)

\(\Leftrightarrow\dfrac{x-5}{2010}-1+\dfrac{x-4}{2011}-1=\dfrac{x-2010}{5}-1+\dfrac{x-2011}{4}-1\)

\(\Leftrightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}=\dfrac{x-2015}{5}+\dfrac{x-2015}{4}\)

\(\Rightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}-\dfrac{x-2015}{5}-\dfrac{x-2015}{4}=0\)

\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x-2015=0\)

=> x=2015

Vậy phương trình có nghiệm x=2015