Đáp án:

Giải thích các bước giải:

Bài 3:
$a) \dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-(\sqrt{x}-\sqrt{y})^{2}\\=\dfrac{(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+\sqrt{y})}{\sqrt{x}+\sqrt{y}}\\=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\\=\sqrt{xy}\\b) \sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\\=\sqrt{\dfrac{x-2\sqrt{x}+1}{\sqrt{x+2\sqrt{x}+1}}}\\=\dfrac{\sqrt{(\sqrt{x}-1)^{2}}}{\sqrt{(\sqrt{x}+1)^{2}}}\\=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}.\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\\=\dfrac{(\sqrt{x}-1)(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}\\=\dfrac{(\sqrt{x}-1)^{2}}{x-1}\\=\dfrac{x-2\sqrt{x}+1}{x-1}\\c) \dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{(y-2\sqrt{y}+1)^{2}}{(x-1)^{4}}}\\=\dfrac{x-1}{\sqrt{y}-1}.\dfrac{y-2\sqrt{y}+1}{(x-1)^{2}}\\=\dfrac{1}{\sqrt{y}-1}.\dfrac{(\sqrt{y}-1)^{2}}{x-1}\\=\dfrac{\sqrt{y}-1}{x-1}$