Question

[Kim Khánh Linh]

Bài 54 (trang 30 SGK Toán 9 Tập 1)

Rút gọn biểu thức sau (giả thiết các biểu thức chữ đều có nghĩa):

$\dfrac{2+\sqrt{2}}{1+\sqrt{2}}$ ; $\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}$ ;  $\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}$ ; $\dfrac{a-\sqrt{a}}{1-\sqrt{a}}$ ; $\dfrac{p-2 \sqrt{p}}{\sqrt{p}-2}$.

Answer 1

[Phạm Bá Huy]

a) $(\sqrt{a}+1)(b\sqrt{a}+1)$.
b) $(x-y)(\sqrt{x}+\sqrt{y})$.

Answer 2

[Đỗ Văn Công]

\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)

\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)

\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)

\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)