Đáp án:\(Q=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
Giải thích các bước giải:
\(Q=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)(x \ge 0,x \ne 4)\)
\(Q=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{(\sqrt{x}-2)(\sqrt{x}+2)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
\(Q=\left(\dfrac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{7}{(\sqrt{x}-2)(\sqrt{x}+2)}\right):\left(\dfrac{(\sqrt{x}-1)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}\right)\)
\(Q=\left(\dfrac{\sqrt{x}-2+7}{(\sqrt{x}-2)(\sqrt{x}+2)}\right):\dfrac{x+\sqrt{x}-2-x+4}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(Q=\dfrac{\sqrt{x}+5}{(\sqrt{x}-2)(\sqrt{x}+2)}:\dfrac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(Q=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
