a) xy2 + 2xy - 243y + x = 0
\(\Leftrightarrow\)x ( y + 1 )2 = 243y
Mà ( y ; y + 1 ) = 1 nên 243 \(⋮\)( y + 1 )2
Mặt khác ( y + 1 ) 2 là số chính phương nên ( y + 1 )2 \(\in\){ 32 ; 92 }
+) ( y + 1 )2 = 32 \(\Rightarrow\orbr{\begin{cases}y+1=3\\y+1=-3\end{cases}\Rightarrow\orbr{\begin{cases}y=2\Rightarrow x=54\\y=-4\Rightarrow x=-108\end{cases}}}\)
+) ( y + 1 )2 = 92 \(\Rightarrow\orbr{\begin{cases}y+1=9\\y+1=-9\end{cases}\Rightarrow\orbr{\begin{cases}y=8\Rightarrow x=24\\y=-10\Rightarrow x=-30\end{cases}}}\)
vậy ...
b) \(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\)( đk : x > 0 )
\(\Leftrightarrow\sqrt{x^2+12}-4=3x+\sqrt{x^2+5}-9\)
\(\Leftrightarrow\sqrt{x^2+12}-4=3x-6+\sqrt{x^2+5}-3\)
\(\Leftrightarrow\frac{x^2-4}{\sqrt{x^2+12}+4}=3\left(x-2\right)+\frac{x^2-4}{\sqrt{x^2+5}+3}\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{\sqrt{x^2+12}+4}-\frac{x+2}{\sqrt{x^2+5}+3}-3\right)=0\)
Vì \(\sqrt{x^2+12}+4>\sqrt{x^2+5}+3\Rightarrow\frac{x+2}{\sqrt{x^2+12}+4}< \frac{x+2}{\sqrt{x^2+5}+3}\)
Do đó : \(\frac{x+2}{\sqrt{x^2+12}+4}-\frac{x+2}{\sqrt{x^2+5}+3}-3< 0\)nên x - 2 = 0 \(\Leftrightarrow\)x = 2
