Question

BT: giải các pt sau

1. \(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)

2. \(\frac{3x+2}{4}+\frac{x+3}{2}=\frac{x-1}{3}-\frac{-x-1}{12}\)

3.\(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2+3}{1-x^2}=0\)

4. \(\frac{1}{x+2}-\frac{3x}{x-2}=\frac{16}{x^2-4}\)

Answer 1

1.\(\frac{x+1}{2013}\)+\(\frac{x+2}{2012}\)=\(\frac{x+3}{2011}\)+\(\frac{x+4}{2010}\)

⇔\(\frac{x+1}{2013}\)+1+\(\frac{x+2}{2012}\)+1=\(\frac{x+3}{2011}\)+1+\(\frac{x+4}{2010}\)+1

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)=\(\frac{x+2014}{2011}\)+\(\frac{x+2014}{2010}\)

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)-\(\frac{x+2014}{2011}\)-\(\frac{x+2014}{2010}\)=0

⇔(x+2014)(\(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\))=0

Mà \(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\)≠0

⇔x+2014=0

⇔x=-2014

Vậy tập nghiệm của phương trình đã cho là:S={-2014}

Answer 2

2.\(\frac{3x+2}{4}\)+\(\frac{x+3}{2}\)=\(\frac{x-1}{3}\)-\(\frac{-x-1}{12}\)

⇔\(\frac{3\left(3x+2\right)}{12}\)+\(\frac{6\left(x+3\right)}{12}\)=\(\frac{4\left(x-1\right)}{12}\)+\(\frac{x+1}{12}\)

⇒9x+6+6x+18=4x-4+x+1

⇒15x+24=5x-3

⇒15x-5x=-3-24

⇒10x=-27

⇒ x=-\(\frac{27}{10}\)

Vậy tập nghiệm của phương trình đã cho là S={-\(\frac{27}{10}\)}

Answer 3

\(3.\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2+3}{1-x^2}=0ĐKXĐ:x\ne\pm1\)

\(\frac{1+x}{x-1}-\frac{x-1}{1+x}+\frac{x^2+3}{\left(1+x\right)\left(1-x\right)}=0\)

\(-3+7x-5x^2+x^3=0\)

\(\left(x-3\right)\left(x-1\right)\left(x-1\right)=0\)

\(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)Theo ĐKXĐ => x=3

Answer 4

\(4.\frac{1}{x+2}-\frac{3x}{x-2}=\frac{16}{x^2-4}ĐKXĐ:x\ne\pm2\)

\(\frac{1}{x+2}-\frac{3x}{x-2}=\frac{16}{x^2-2^2}\)

\(\frac{1}{x+2}-\frac{3x}{x-2}=\frac{16}{\left(x+2\right)\left(x-2\right)}\)

\(-5x-2-3x^2=16\)

\(-5x-2-3x^2-16=0\)

\(5x+18+3x^2=0\)

\(\left[{}\begin{matrix}x=\frac{-5+\sqrt{191t}}{6}\\x=\frac{-5-\sqrt{191t}}{6}\end{matrix}\right.\)