a) Ta có:
\(x=11-4\sqrt{7}=\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2=\left(\sqrt{7}-2\right)^2\)
Thay vào P ta có:
\(P=\dfrac{\sqrt{\left(\sqrt{7}-2\right)^2}-2}{\sqrt{\left(\sqrt{7}-2\right)^2}}=\dfrac{\sqrt{7}-2-2}{\sqrt{7}-2}=\dfrac{\sqrt{7}-4}{\sqrt{7}-2}\)
b) \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}+\dfrac{7\sqrt{x}-3}{x-9}-\dfrac{3\sqrt{x}-x}{\sqrt{x}-3}\)
\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\)
\(Q=\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)
\(Q=\dfrac{x-4\sqrt{x}+3+7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)
\(Q=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)
\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(Q=\dfrac{\sqrt{x}+x-3\sqrt{x}}{\sqrt{x}-3}\)
\(Q=\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\)
c) \(M=P\cdot Q\)
\(M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\)
\(M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)
\(M=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-3}\)
Mà: \(M\ge0\) khi
\(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-3}\ge0\) ta có: \(\left(\sqrt{x}-2\right)^2\ge0\forall x\)
\(\Rightarrow\sqrt{x}-3>0\)
\(\Rightarrow\sqrt{x}>3\)
\(\Rightarrow x>9\)
