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`

nhưng đề bài là j bạn ơi

Đề là gì vậy

ko hiểu đề

b: I=-15+13=-2

Bài làm

\(A=\frac{19}{5}xy^2.\left(x^3y\right).\left(-3x^{13}y^5\right)^0\)

\(A=\frac{19}{5}xy^2.\left(x^3y\right).1\)

\(A=\frac{19}{5}xy^2.\left(x^3y\right)\)

\(A=\frac{19}{5}x^4y^3\)

Vậy \(A=\frac{19}{5}x^4y^3\)

\(A=\frac{19}{5}xy^2\left(x^3y\right)\left(-3x^{13}y^5\right)^0\)

\(=\frac{19}{5}xy^2\left(x^3y\right)\)

\(=\frac{19}{5}\left(xx^3\right)\left(y^2y\right)\)

\(=\frac{19}{5}x^4y^3\)

\(\dfrac{3}{4}=4x-\dfrac{2}{8}\)

\(\Leftrightarrow4x=\dfrac{3}{4}+\dfrac{2}{8}\)

\(\Leftrightarrow4x=\dfrac{3}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow4x=1\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

played

is riding

will be

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Does..........do