Question

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Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

a: Thay x=4/9 vào A, ta được:

\(A=\dfrac{\sqrt{\dfrac{4}{9}}-3}{\dfrac{4}{9}+7}=\dfrac{\dfrac{2}{3}-3}{\dfrac{67}{9}}\)

\(=-\dfrac{7}{3}\cdot\dfrac{9}{67}=\dfrac{-7\cdot3}{67}=-\dfrac{21}{67}\)

b: \(B=\left(\dfrac{x+\sqrt{x}+10}{x-9}-\dfrac{1}{\sqrt{x}-3}\right):\dfrac{1}{\sqrt{x}-3}\)

\(=\left(\dfrac{x+\sqrt{x}+10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{\sqrt{x}-3}{1}\)

\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{1}\)

\(=\dfrac{x+7}{\sqrt{x}+3}\)

c: Q=A*B

\(=\dfrac{\sqrt{x}-3}{x+7}\cdot\dfrac{x+7}{\sqrt{x}+3}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

Để |Q|>Q thì Q<0

=>\(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}< 0\)

=>\(\sqrt{x}-3< 0\)

=>\(\sqrt{x}< 3\)

=>0<=x<9