a) \(A=\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}+2}-\dfrac{8\sqrt[]{x}}{4-x}\right):\dfrac{\sqrt[]{x}+2}{1-2\sqrt[]{x}}\)
\(A=\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}+2}+\dfrac{8\sqrt[]{x}}{x-4}\right):\dfrac{\sqrt[]{x}+2}{1-2\sqrt[]{x}}\)
\(A=\left[\dfrac{\left(\sqrt[]{x}-2\right)^2}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}+\dfrac{8\sqrt[]{x}}{\left(\sqrt[]{x}-2\right)\left(\sqrt[]{x}+2\right)}\right]:\dfrac{\sqrt[]{x}+2}{1-2\sqrt[]{x}}\)
\(A=\dfrac{\left(\sqrt[]{x}-2\right)^2+8\sqrt[]{x}}{\left(\sqrt[]{x}-2\right)\left(\sqrt[]{x}+2\right)}.\dfrac{1-2\sqrt[]{x}}{\sqrt[]{x}+2}\)
\(A=\dfrac{x-4\sqrt[]{x}+4+8\sqrt[]{x}}{\left(\sqrt[]{x}-2\right)\left(\sqrt[]{x}+2\right)}.\dfrac{1-2\sqrt[]{x}}{\sqrt[]{x}+2}\)
\(A=\dfrac{x+4\sqrt[]{x}+4}{\left(\sqrt[]{x}-2\right)\left(\sqrt[]{x}+2\right)}.\dfrac{1-2\sqrt[]{x}}{\sqrt[]{x}+2}\)
\(A=\dfrac{\left(\sqrt[]{x}+2\right)^2.\left(1-2\sqrt[]{x}\right)}{\left(\sqrt[]{x}-2\right)\left(\sqrt[]{x}+2\right)^2}\)
\(A=\dfrac{1-2\sqrt[]{x}}{\sqrt[]{x}-2}\)
b) Để \(A=-\dfrac{1}{3}\) ⇔ \(\dfrac{1-2\sqrt[]{x}}{\sqrt[]{x}-2}=-\dfrac{1}{3}\)
⇔ \(3\left(1-2\sqrt[]{x}\right)=-\left(\sqrt[]{x}-2\right)\)
⇔ \(3-6\sqrt[]{x}=-\sqrt[]{x}+2\)
⇔ \(-\sqrt[]{x}+6\sqrt[]{x}=3-2\)
⇔ \(5\sqrt[]{x}=1\)
⇔ \(\sqrt[]{x}=\dfrac{1}{5}\)
⇔ \(x=\dfrac{1}{25}\) (bình phương 2 vế)
Sau khi thử nghiệm lại, ta thấy nghiệm \(x=\dfrac{1}{25}\) hợp lý
Vậy để \(A=-\dfrac{1}{3}\) thì \(x=\dfrac{1}{25}\)
Chúc bn học tốt:)))
