Question

giải phương trình sau:

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

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

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

Answer 1

ai bít thì giúp mình với nhé

Answer 2

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

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

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

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

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

mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)

\(\Leftrightarrow2015-x=0\)

\(\Leftrightarrow x=2015\)

KL : PT có nghiệm \(S=\left\{2015\right\}\)

Answer 3

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

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

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

\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}-\frac{2015-x}{2002}-\frac{2015-x}{2003}=0\)

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

Vì \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\ne0\)

\(\Leftrightarrow2015-x=0\)

\(\Leftrightarrow x=2015\)

Vậy x=2015

Answer 4

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

\(\Rightarrow\left(\frac{15-x}{2000}+1\right)+\left(\frac{14-x}{2001}+1\right)=\)\(\left(\frac{13-x}{2002}+1\right)+\left(\frac{14-x}{2003}+1\right)\)   \(\Rightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015-x}{2002}+\frac{2015-x}{2003}\)

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

\(\Rightarrow\left(2015-x\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Rightarrow2015-x=0\)( vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\))

 \(\Rightarrow x=2015\)

Vậy x =2015

b) tương tự

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

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

 \(\Leftrightarrow\left(x^2+1\right).\left[\left(x^2+1\right)+x\right]+2x.\left[\left(x^2+1\right)+x\right]=0\)

 \(\Leftrightarrow\left[\left(x^2+1\right)+x\right].\left[x^2+1+2x\right]=0\)

 \(\Leftrightarrow\left(x^2+x+1\right).\left(x^2+2x+1\right)=0\)

  \(\Leftrightarrow\left(x+1\right)^2.\left(x+1\right)^2=0\)

 \(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\)\(\Leftrightarrow x=1\)

Vậy x=1