\(a,P=\dfrac{7\sqrt{x}+3+x-3\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\\ b,x=\dfrac{4}{25}\Leftrightarrow\sqrt{x}=\dfrac{2}{5}\\ \Leftrightarrow P=\dfrac{\dfrac{2}{5}}{\dfrac{2}{5}+3}=\dfrac{\dfrac{2}{5}}{\dfrac{17}{5}}=\dfrac{2}{17}\\ c,P>0\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+3}>0\)
Ta có \(\sqrt{x}\ge0,\forall x;\sqrt{x}+3\ge3>0,\forall x\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+3}\ge0,\forall x\)
Vậy P luôn lớn hơn 0 với mọi x
a) \(P=\dfrac{7\sqrt{x}+3+\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
b) \(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{\sqrt{\dfrac{4}{25}}}{\sqrt{\dfrac{4}{25}}-3}=\dfrac{\dfrac{2}{5}}{\dfrac{2}{5}-3}=-\dfrac{2}{13}\)
c)\(\dfrac{\sqrt{x}}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow\sqrt{x}< 3\)
Kết hợp ĐK:
\(\Rightarrow0\le x< 9\)
ở trên sai hết nha 😭
\(a,P=\dfrac{7\sqrt{x}+3+x-3\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(b,x=\dfrac{4}{25}\Leftrightarrow\sqrt{x}=\dfrac{2}{5}\Leftrightarrow P=\dfrac{\dfrac{2}{5}}{\dfrac{2}{5}-3}=\dfrac{2}{5}:\left(-\dfrac{13}{5}\right)=-\dfrac{2}{13}\)
\(c,P< 0\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}\ge0;\sqrt{x}-3\ne0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
