Rút gọn !!!!!!!!!!!!!!!!!!!!!!

Đáp án:

$B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-8}{2\sqrt{x}-x}=\dfrac{\sqrt{x}-4}{\sqrt{x}}$

Giải thích các bước giải:

$B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-8}{2\sqrt{x}-x}\,\,\,(x>0,x\ne4,x\ne16)\\=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{x-2\sqrt{x}}\\=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{\sqrt{x}(x-2)}\\=\dfrac{\sqrt{x}(\sqrt{x}-1)-(5\sqrt{x}-8)}{\sqrt{x}(\sqrt{x}-2)}\\=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}(\sqrt{x}-2)}\\=\dfrac{x-6\sqrt{x}+8}{\sqrt{x}(\sqrt{x}-2)}\\=\dfrac{(\sqrt{x}-2)(\sqrt{x}-4)}{\sqrt{x}(\sqrt{x}-2)}\\=\dfrac{\sqrt{x}-4}{\sqrt{x}}$