a: Thay x=9 vào A, ta được:\(A=\dfrac{4\left(3+1\right)}{25-9}=\dfrac{4\cdot4}{16}=1\)b: \(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)\(=\left(\dfrac{15-\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{2}{\sqrt{x}+5}\right)\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)\(=\dfrac{15-\sqrt{x}+2\left(\sqrt[]{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)c: \(P=A:B=\dfrac{4\left(\sqrt{x}+1\right)}{25-x}:\dfrac{1}{\sqrt{x}+1}=\dfrac{4}{25-x}\)Để P là số nguyên lớn nhất thì 25-x=1=>x=24Câu trả lời: a: Thay x=9 vào A, ta được:\(A=\dfrac{4\left(3+1\right)}{25-9}=\dfrac{4\cdot4}{16}=1\)b: \(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)\(=\left(\dfrac{15-\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{2}{\sqrt{x}+5}\right)\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)\(=\dfrac{15-\sqrt{x}+2\left(\sqrt[]{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)c: \(P=A:B=\dfrac{4\left(\sqrt{x}+1\right)}{25-x}:\dfrac{1}{\sqrt{x}+1}=\dfrac{4}{25-x}\)Để P là số nguyên lớn nhất thì 25-x=1=>x=24