a.Thay `x=25` vào `A`, ta được:
\(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{25}-2}{\sqrt{25}+1}=\dfrac{3}{6}=\dfrac{1}{2}\)
b.
\(B=\dfrac{4}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{2-\sqrt{x}}+\dfrac{12}{x-\sqrt{x}-2}\)
\(B=\dfrac{4}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{2-\sqrt{x}}+\dfrac{12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{4\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{4\sqrt{x}-8-\left(x-1\right)+12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{-x+4\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\left(5-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{5-\sqrt{x}}{\sqrt{x}-2}\)
c.
\(P=A.B\)
\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}.\dfrac{5-\sqrt{x}}{\sqrt{x}-2}\)
\(P=\dfrac{5-\sqrt{x}}{\sqrt{x}+1}\)
\(P=-\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(P=-1+\dfrac{6}{\sqrt{x}+1}\)
Để `P` nguyên thì \(\dfrac{6}{\sqrt{x}+1}\in Z\) hay \(\sqrt{x}+1\in U\left(6\right)=\left\{\pm1;\pm3;\pm6\right\}\)
`@`\(\sqrt{x}+1=1\rightarrow x=0\)
`@`\(\sqrt{x}+1=-1\rightarrow\) vô lý
`@`\(\sqrt{x}+1=3\rightarrow x=4\)
`@`\(\sqrt{x}+1=-3\rightarrow\) vô lý
`@`\(\sqrt{x}+1=6\rightarrow x=25\)
`@`\(\sqrt{x}+1=-6\rightarrow\) vô lý
Vậy \(x\in\left\{0;4;25\right\}\) thì `P` nhân giá trị nguyên
