trục căn thức ở mẫu vs giả thiết các biểu thức chữ đều có nghĩa ( từ bài 50 đến bài 52)
BT50
\(\dfrac{5}{\sqrt{10}}\);\(\dfrac{5}{2\sqrt{5}}\);\(\dfrac{1}{3\sqrt{20}}\);\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}\);\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}\)
BT51
\(\dfrac{3}{\sqrt{3}+1}\);\(\dfrac{2}{\sqrt{3}-1}\);\(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\);\(\dfrac{b}{3+\sqrt{b}}\);\(\dfrac{p}{2\sqrt{p}-1}\)
BT52
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}\);\(\dfrac{3}{\sqrt{10}+\sqrt{7}}\);\(\dfrac{1}{\sqrt{x}-\sqrt{y}}\);\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}\)
Bài 50:
\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\sqrt{20}}=\dfrac{1}{6\sqrt{5}}=\dfrac{\sqrt{5}}{30}\)
\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
