Cho [tex]\sqrt{x}+\sqrt{3x-2}= x^{2}+1[/tex]
Tìm x
Đk:[tex]x\geq \frac{3}{2 }[/tex]
Đặt [tex]\left\{\begin{matrix} \sqrt{x}=a(a\geq \sqrt{\frac{3}{2 }} )\\ \sqrt{3x-2}=b(b\geq 0) \end{matrix}\right.[/tex]
Khi đó ta có hệ [tex]\left\{\begin{matrix} a+b=a^4+1\\ 3a^2-b^2=2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} b=a^4-a+1\\ 3a^2-(a^4-a+1)^2=2\end{matrix}\right.[/tex]
Giải [tex]3a^2-(a^4-a+1)^2=2\Leftrightarrow (a-1)[3a+3-(a^2+a+1)(a^5-a^2+2a)]=0[/tex]
Vì [tex]3a+3-(a^2+a+1)(a^5-a^2+2a)< 0;\forall a\geq \sqrt{\frac{3}{2}}[/tex]
[tex]\Rightarrow a=1\Rightarrow x=1[/tex]