Question

giải các phương trình sau

d) 2-x/2001 - 1 = 1-x/2002 - x/2003

e) 150-x/25  + 188-x/21 + 201-x/19 +171-x/23 =0

Answer 1

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+\left(\frac{x}{2003}-1\right)\)

\(\Leftrightarrow\frac{2-x+2001}{2001}=\frac{1-x+2002}{2002}+\frac{x-2003}{2003}\)

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

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

\(\Leftrightarrow x-2003=0\)\(\left(v\text{ì}\frac{1}{2003}+\frac{1}{2001}-\frac{1}{2002}\ne0\right)\)

\(\Leftrightarrow x=2003\)

Vậy \(S=\left\{2003\right\}\)

Answer 2

d)Ta có :  \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1-2=\frac{1-x}{2002}+1+1-\frac{x}{2003}-2\)\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

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

\(\Leftrightarrow2003-x=0\Leftrightarrow x=2003\)

Vậy phương trình có tập nghiệm S = { 2003 }