Question

Giải các phương trình sau:

a) \(\dfrac{2x+1}{x-3}-\dfrac{x}{x+3}=0\)

b)\(\left(x-4\right).\left(x-5\right)=12\)

c)\(\dfrac{3-x}{2008}-1=\dfrac{2-x}{2009}-\dfrac{x}{2011}\)

Answer 1

a)\(\dfrac{2x+1}{x-3}-\dfrac{x}{x+3}=0\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(x+3\right)-\left(x-3\right)x=0\)

\(\Leftrightarrow2x^2+7x+3-x^2+3x=0\)

\(\Leftrightarrow x^2+10x+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5+\sqrt{22}\\x=-5-\sqrt{22}\end{matrix}\right.\)(tm)

Answer 2

b.

\(\left(x-4\right)\left(x-5\right)=12\)

\(\Leftrightarrow x^2-9x+20-12-0\)

\(\Leftrightarrow x^2-9x+8=0\)

\(\Leftrightarrow x^2-8x-x+8=0\)

\(\Leftrightarrow x\left(x-8\right)-\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

Vậy........

Answer 3

c)\(\dfrac{3-x}{2008}-1=\dfrac{2-x}{2009}-\dfrac{x}{2011}\)

\(\Leftrightarrow\dfrac{3-x}{2008}+1=\dfrac{2-x}{2009}+1-\dfrac{x}{2011}+1\)

\(\Leftrightarrow\dfrac{2011-x}{2008}=\dfrac{2011-x}{2009}+\dfrac{2011-x}{2011}\)

\(\Leftrightarrow\left(2011-x\right)\left(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2011}\right)=0\)

\(\Leftrightarrow x=2011\)(vì \(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2011}\ne0\))