a
\(B=\left(\dfrac{3+\sqrt{x}}{9-x}-\dfrac{3-\sqrt{x}}{9-x}\right).\dfrac{\sqrt{x}+3}{\sqrt{x}}\\ =\left(\dfrac{3+\sqrt{x}-3+\sqrt{x}}{9-x}\right).\dfrac{\sqrt{x}+3}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\sqrt{x}}\\ =\dfrac{2}{\sqrt{x}-3}\)
b
\(B=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{2}{\sqrt{x}-3}=\dfrac{1}{2}\left(x>3\right)\)
\(\Leftrightarrow\sqrt{x}-3=4\\ \Leftrightarrow\sqrt{x}=7\\ \Leftrightarrow x=49\left(nhận\right)\)
a) \(B=\left(\dfrac{1}{3-\sqrt{x}}-\dfrac{1}{3+\sqrt{x}}\right)\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
\(B=\left(\dfrac{3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}-\dfrac{3-\sqrt{x}}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right)\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
\(B=\left(\dfrac{3+\sqrt{x}-3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right)\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
\(B=\dfrac{2\sqrt{x}}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
\(B=\dfrac{2}{3-\sqrt{x}}\)
b) \(B=\dfrac{1}{2}\) khi
\(\dfrac{2}{3-\sqrt{x}}=\dfrac{1}{2}\)
\(\Leftrightarrow4=3-\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=3-4\)
\(\Leftrightarrow\sqrt{x}=-1\) (vô lý)
Không có x thỏa mãn
giúp e vs ạ ;-;
giúp e vs ạ :<