Đáp án:
a)
$x_1=\dfrac{\sqrt{5}+1}{2}$
$x_2=\dfrac{\sqrt{5}-1}{2}$
b)
$A=1$
Giải thích các bước giải:
a)
$x^2-\sqrt{5}x+1=0$
$\Delta=(-\sqrt{5})^2-4.1.1=5-4=1>0$
$\to$ Phương trình có 2 nghiệm phân biệt
$x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{\sqrt{5}+1}{2}$
$x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{\sqrt{5}-1}{2}$
b)
$A=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):(x-y)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$(x,y>0)$
$A=\left(\dfrac{(\sqrt{x})^3-(\sqrt{y})^3}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):(x-y)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=\left(\dfrac{(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+y)}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):(x-y)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=\dfrac{x-\sqrt{xy}+y-\sqrt{xy}}{x-y}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=\dfrac{x-2\sqrt{xy}+y}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=\dfrac{(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=\dfrac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
$A=1$
