Bài 51 trang 30 SGK Toán 9 tập 1

Đề bài

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

\(\dfrac{3}{\sqrt{3}+1};\,\,\,\dfrac{2}{\sqrt{3}-1};\,\,\,\dfrac{2+\sqrt{3}}{2-\sqrt{3}};\,\,\,\dfrac{b}{3+\sqrt{b}};\,\,\,\dfrac{p}{2\sqrt{p}-1}.\)

Lời giải chi tiết

+ Ta có:

\(\dfrac{3}{\sqrt{3}+1}=\dfrac{3(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\dfrac{3\sqrt 3 - 3.1}{(\sqrt 3)^2-1^2}\)

\(=\dfrac{3\sqrt 3 -3}{3-1}=\dfrac{3\sqrt{3}-3}{2}\).

+ Ta có:

\(\dfrac{2}{\sqrt{3}-1}=\dfrac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\dfrac{2(\sqrt 3 + 1)}{(\sqrt 3)^2-1^2}\)

\(=\dfrac{2(\sqrt 3 + 1)}{3-1}=\dfrac{2(\sqrt{3}+1)}{2}=\sqrt{3}+1\).

+ Ta có:

\(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{(2+\sqrt{3}).(2+\sqrt 3)}{(2-\sqrt{3})(2+\sqrt{3})}=\dfrac{(2+\sqrt{3})^2}{2^2-(\sqrt{3})^2}\)

\(=\dfrac{2^2+2.2.\sqrt 3+(\sqrt{3})^2}{4-3}\)\(=\dfrac{4+4\sqrt 3+3}{1}=\dfrac{(4+3)+4\sqrt 3}{1}\)

\(=\dfrac{7+4\sqrt 3}{1}=7+4\sqrt{3}\).

+ Ta có:

\(\dfrac{b}{3+\sqrt{b}}=\dfrac{b(3-\sqrt{b})}{(3+\sqrt{b})(3-\sqrt{b})}\)

\(=\dfrac{b(3-\sqrt{b})}{3^2-(\sqrt b)^2}=\dfrac{b(3-\sqrt{b})}{9-b};(b\neq 9)\).

+ Ta có:

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

\(=\dfrac{p(2\sqrt{p}+1)}{(2\sqrt{p})^2-1^2}=\dfrac{p(2\sqrt{p}+1)}{4p-1}\)