Question

[Hoàng Trần Đình]

Giải giúp mình câu 5,6,7,8,9,10

 

Answer 1

[Nguyễn Lê Phước Thịnh]

9:

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

a: Khi x=25 thì \(A=\dfrac{7}{5+8}=\dfrac{7}{13}\)

b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

c: Để B là số nguyên thì \(\sqrt{x}+8⋮\sqrt{x}+3\)

=>\(\sqrt{x}+3+5⋮\sqrt{x}+3\)

=>\(\sqrt{x}+3\inƯ\left(5\right)\)

=>\(\sqrt{x}+3\in\left\{1;-1;5;-5\right\}\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

d: \(P=A\cdot B=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\cdot\dfrac{7}{\sqrt{x}+8}=\dfrac{7}{\sqrt{x}+3}\)

Để P là số nguyên thì \(7⋮\sqrt{x}+3\)

mà \(\sqrt{x}+3>=3\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+3=7\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)