$2x\sqrt{x^2-x+1}+4\sqrt{3x+1}=2x^2+2x+6$
Edited by Dinh Xuan Hung, 17-06-2015 - 17:23.
$2x\sqrt{x^2-x+1}+4\sqrt{3x+1}=2x^2+2x+6$
Edited by Dinh Xuan Hung, 17-06-2015 - 17:23.
$\left\{\begin{matrix} \sqrt{5-\sqrt{y+5}}=x & \\ 2x^{3}=y(y^2+x^2)& \end{matrix}\right.$
$2x\sqrt{x^2-x+1}+4\sqrt{3x+1}=2x^2+2x+6$
PT $\Leftrightarrow (x^2-2.x.\sqrt{x^2-x+1}+x^2-x+1)+(3x+1-4.\sqrt{3x+1}+4)=0$
$\Leftrightarrow (x-\sqrt{x^2-x+1})^2+(\sqrt{3x+1}-2)^2=0$
Bạn tự giải tiếp
$\left\{\begin{matrix} \sqrt{5-\sqrt{y+5}}=x (1) & \\ 2x^{3}=y(y^2+x^2) (2)& \end{matrix}\right.$
PT $(2)\Leftrightarrow 2x^3-x^2y-y^3=0$
$\Leftrightarrow x^3-x^2y+x^3-y^3=0$
$\Leftrightarrow x^2(x-y)+(x-y)(x^2+xy+y^2)=0$
$\Leftrightarrow (x-y)(2x^2+xy+y^2)$
$\Leftrightarrow x=y\vee x=y=0$ vì dễ dàng chỉ ra:$2x^2+xy+y^2\geq 0$. Đẳng thức xảy ra tại $x=y=0$
$\Leftrightarrow x=y$
Thay vào PT $(1)$ giải bằng cách bình phương lên.
Edited by vda2000, 17-06-2015 - 13:36.
$\boxed{\textrm{Silence is the peak of contempt!}}$
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