Bài 50 trang 30 SGK Toán 9 tập 1

Đề bài

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

\(\dfrac{5}{\sqrt{10}};\,\,\, \dfrac{5}{2\sqrt{5}};\,\,\, \dfrac{1}{3\sqrt{20}};\,\,\, \dfrac{2\sqrt{2}+2}{5\sqrt{2}};\,\,\, \dfrac{y+b.\sqrt{y}}{b. \sqrt{y}}.\) 

Lời giải chi tiết

+ Ta có: 

\(\dfrac{5}{\sqrt{10}}=\dfrac{5.\sqrt{10}}{\sqrt{10}.\sqrt{10}}=\dfrac{5\sqrt{10}}{(\sqrt{10})^2}=\dfrac{5\sqrt{10}}{10}\)

\(=\dfrac{5.\sqrt{10}}{5.2}\)\(=\dfrac{\sqrt{10}}{2}\).

+ Cách 1: Ta có:

\(\dfrac{5}{2\sqrt{5}}=\dfrac{5.\sqrt 5}{2\sqrt 5.\sqrt 5}=\dfrac{5\sqrt{5}}{2.(\sqrt 5.\sqrt 5)}=\dfrac{5\sqrt{5}}{2(\sqrt 5)^2}\)

\(=\dfrac{5\sqrt 5}{2.5}=\dfrac{\sqrt 5}{2}\).

Cách 2: Ta có: 

\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt 5.\sqrt 5}{2.\sqrt 5}=\dfrac{\sqrt 5}{2}\).

+ Ta có:

\(\dfrac{1}{3\sqrt{20}}=\dfrac{1.\sqrt{20}}{3\sqrt{20}.\sqrt{20}}=\dfrac{\sqrt{20}}{3.(\sqrt{20}.\sqrt{20})}=\dfrac{\sqrt{20}}{3.(\sqrt{20})^2}\)

              \(=\dfrac{\sqrt{20}}{3.20}=\dfrac{\sqrt{2^2.5}}{60}=\dfrac{2\sqrt 5}{60}=\dfrac{2\sqrt 5}{2.30}=\dfrac{\sqrt 5}{30}\).

+ Cách 1: Ta có:

\(\dfrac{(2\sqrt{2}+2)}{5.\sqrt 2}=\dfrac{(2\sqrt 2+2).\sqrt 2}{5\sqrt 2. \sqrt 2}=\dfrac{2\sqrt 2.\sqrt 2+2.\sqrt 2}{5.(\sqrt 2)^2}\\=\dfrac{2.2+2\sqrt 2}{5.2}=\dfrac{2(2+\sqrt 2)}{5.2}=\dfrac{2+\sqrt 2}{5}\).

Cách 2: Ta có:

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

+ Cách 1: Ta có:

 \(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{(y+b\sqrt y).\sqrt y}{b\sqrt y .\sqrt y}=\dfrac{y\sqrt y+b\sqrt y.\sqrt y}{b.(\sqrt y)^2}\)

                    \(= \dfrac{y\sqrt y+b(\sqrt y)^2}{by}=\dfrac{y\sqrt y+by}{by}\)

                    \(=\dfrac{y(\sqrt y+b)}{b.y}=\dfrac{\sqrt y+b}{b}\).

Cách 2: Ta có:

\(\dfrac{{y + b\sqrt y }}{{b\sqrt y }} = \dfrac{{{{\left( {\sqrt y } \right)}^2} + b\sqrt y }}{{b\sqrt y }} \)\(= \dfrac{{\sqrt y \left( {\sqrt y  + b} \right)}}{{b\sqrt y }} = \dfrac{{\sqrt y  + b}}{b}\)