Question

giải pt sau

Answer 1

\(\Leftrightarrow5+\left(x+3\right)\left(x-3\right)=0\\ \Leftrightarrow5+x^2-9=0\\ \Leftrightarrow x^2-4=0\Rightarrow x^2=4\Rightarrow x=\pm2\)

Answer 2

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Answer 3

\(ĐK:x\ne2,3\)

\(\Leftrightarrow\dfrac{5}{\left(x-3\right)\left(2-x\right)}+\dfrac{x+3}{2-x}=0\)

\(\Leftrightarrow\dfrac{5+\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(2-x\right)}=0\)

\(\Leftrightarrow5+x^2-9=0\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)