Question

Answer 1

đk  x ≠ 4 ; x ≥ 0

\(a,\\P=\left(\dfrac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{4-x}\right).\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\\ =\dfrac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{2+\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ =\dfrac{8\sqrt{x}+4x}{2+\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ =\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{2+\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ =\dfrac{4x}{\sqrt{x}-3}\)

\(ĐêP=1\\ \dfrac{4x}{\sqrt{x}-3}=1\\ 4x=\sqrt{x}-3\\ 4x-\sqrt{x}+3=0\)

=> pt vô no

Vậy ko có giá trị nào của x để P =1