Bài 52 trang 30 SGK Toán 9 tập 1

Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa:

\(\dfrac{2}{\sqrt{6}-\sqrt{5}};\,\,\ \dfrac{3}{\sqrt{10}+\sqrt{7}};\,\,\, \dfrac{1}{\sqrt{x}-\sqrt{y}};\,\,\, \dfrac{2ab}{\sqrt{a}-\sqrt{b}}\).

Lời giải

+ Ta có:

\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}\)

                   \(=\dfrac{2(\sqrt{6}+\sqrt{5})}{(\sqrt{6})^2-(\sqrt{5})^2}=\dfrac{2(\sqrt{6}+\sqrt{5})}{6-5}\)

                   \(=\dfrac{2(\sqrt{6}+\sqrt{5})}{1}=2(\sqrt{6}+\sqrt{5})\).

+ Ta có:

\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3(\sqrt{10}-\sqrt{7})}{(\sqrt{10}+\sqrt{7})(\sqrt{10}-\sqrt{7})}\)

                    \(=\dfrac{3(\sqrt{10}-\sqrt{7})}{(\sqrt{10})^2-(\sqrt{7})^2}\)\(=\dfrac{3(\sqrt{10}-\sqrt{7})}{10-7}\)

                    \(=\dfrac{3(\sqrt{10}-\sqrt{7})}{3}=\sqrt{10}-\sqrt{7}\).

+ Ta có:

\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\)

\(=\dfrac{\sqrt x + \sqrt y}{(\sqrt x)^2-(\sqrt y)^2}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)

+ Ta có:

\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}\)

\(=\dfrac{2ab(\sqrt a+ \sqrt b)}{(\sqrt a)^2-(\sqrt b)^2}=\dfrac{2ab(\sqrt{a}+\sqrt{b})}{a-b}\).


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