1: \(P=\dfrac{x+3\sqrt{x}+2\sqrt{x}-24}{x-9}=\dfrac{x+5\sqrt{x}-24}{x-9}\)
\(=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
2: \(Q=\dfrac{7}{\sqrt{x}+8}\cdot\dfrac{\sqrt{x}+8}{\sqrt{x}+3}=\dfrac{7}{\sqrt{x}+3}\)
Để Q là số nguyên thì \(\sqrt{x}+3\inƯ\left(7\right)\)
\(\Leftrightarrow\sqrt{x}+3=7\)
hay x=16
`1)` Với `x >= 0,x \ne 9` có:
`P=\sqrt{x}/[\sqrt{x}-3]+[2\sqrt{x}-24]/[x-9]`
`P=[\sqrt{x}(\sqrt{x}+3)+2\sqrt{x}-24]/[(\sqrt{x}-3)(\sqrt{x}+3)]`
`P=[x+3\sqrt{x}+2\srqt{x}-24]/[(\sqrt{x}-3)(\sqrt{x}+3)]`
`P=[(\sqrt{x}-3)(\sqrt{x}+8)]/[(\sqrt{x}-3)(\sqrt{x}+3)]`
`P=[\sqrt{x}+8]/[\sqrt{x}+3]`
`2)` Với `x >= 0,x \ne 9` có:
`Q=7/[\sqrt{x}+8] .P=7/[\sqrt{x}+8] . [\sqrt{x}+8]/[\sqrt{x}+3] =7/[\sqrt{x}+3]`
Để `Q in ZZ<=>7/[\sqrt{x}+3] in ZZ`
`=>\sqrt{x}+3 in Ư_7`
Mà `Ư_7={+-1;+-7}`
\begin{array}{|c|c|c|}\hline \sqrt{x}+3&1&-1&7&-7\\\hline \sqrt{x}&-2(L)&-4(L)&4&-10(L)\\\hline x& & &16(t/m)&\\\hline\end{array}
Vậy `x=16`
