3.
a, ĐK: \(x\ge0;x\ne9\)
\(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\sqrt{x}+3}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
b, \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{5+2}{5+3}=\dfrac{7}{8}\)
4.
a, ĐK: \(x\ge0;x\ne1\)
\(A=\dfrac{15\sqrt{x}+5}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}+2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+11\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=-\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=-\dfrac{5\sqrt{x}+2}{\sqrt{x}+3}\)
3.
c, \(x=\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{\sqrt{3}-1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2}{3-1}=1\)
\(\Rightarrow P=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)
d, \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{3}{4}\)
\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\left(tm\right)\)
3.
e, \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{5}{7}\)
\(\Leftrightarrow7\sqrt{x}+14=5\sqrt{x}+15\)
\(\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
4.
b, \(A=-\dfrac{5\sqrt{x}+2}{\sqrt{x}+3}=-\dfrac{5.2+2}{2+3}=-\dfrac{12}{5}\)
c, \(A=-\dfrac{5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10\sqrt{x}-4=\sqrt{x}+3\)
\(\Leftrightarrow11\sqrt{x}=-7\)
\(\Leftrightarrow\sqrt{x}=-\dfrac{7}{11}\)
\(\Rightarrow\) vô nghiệm
Vậy không tồn tại giá trị x thỏa mãn để \(A=\dfrac{1}{2}\).
