Bài 1 trang 31 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Tính :

a) \( - \dfrac{1}{2}\sqrt {108}  - \dfrac{1}{{15}}\sqrt {75}  - \dfrac{1}{{22}}\sqrt {363}  + \sqrt {12} \);

b) \(2\sqrt {\dfrac{{27}}{2}}  - \sqrt {\dfrac{{48}}{9}}  - \dfrac{2}{5}\sqrt {\dfrac{{75}}{{18}}} \);

c) \(2y\sqrt {45}  + 3\sqrt {20{y^2}} \);     

d) \(3x\sqrt {72x}  - 9\sqrt {50{x^3}} \) với \(x \ge 0\).

Lời giải chi tiết

\(\begin{array}{l}a)\; - \dfrac{1}{2}\sqrt {108}  - \dfrac{1}{{15}}\sqrt {75}  - \dfrac{1}{{22}}\sqrt {363}  + \sqrt {12} \\ =  - \dfrac{1}{2}\sqrt {{6^2}.3}  - \dfrac{1}{{15}}\sqrt {{5^2}.3}  - \dfrac{1}{{22}}\sqrt {{{11}^2}.3}  + \sqrt {{2^2}.3} \\ =  - \dfrac{1}{2}.6\sqrt 3  - \dfrac{1}{{15}}.5\sqrt 3  - \dfrac{1}{{22}}.11\sqrt 3  + 2\sqrt 3 \\ =  - \dfrac{{11\sqrt 3 }}{6}.\end{array}\)

\(\begin{array}{l}b)\;2\sqrt {\dfrac{{27}}{2}}  - \sqrt {\dfrac{{48}}{9}}  - \dfrac{2}{5}\sqrt {\dfrac{{75}}{{18}}} \\ = 2\sqrt {\dfrac{{{3^2}.3.2}}{{{2^2}}}}  - \sqrt {\dfrac{{{4^2}.3}}{{{3^2}}}}  - \dfrac{2}{5}\sqrt {\dfrac{{{5^2}.3}}{{{3^2}.2}}} \\ = 2.\dfrac{{3\sqrt 3 }}{2} - \dfrac{4}{3}\sqrt 3  - \dfrac{2}{5}.\dfrac{5}{3}\sqrt {\dfrac{3}{2}} \\ = 3\sqrt 3  - \dfrac{{4\sqrt 3 }}{3} - \dfrac{2}{3}\dfrac{{\sqrt 6 }}{2}\\ = \dfrac{{5\sqrt 3 }}{3} - \dfrac{{\sqrt 6 }}{3} = \dfrac{{4\sqrt 3 }}{3}.\end{array}\)

\(\begin{array}{l}c)\;2y\sqrt {45}  + 3\sqrt {20{y^2}} \\ = 2y\sqrt {{3^2}.5}  + 3\sqrt {{2^2}.5} .\sqrt {{y^2}} \\ = 6y\sqrt 5  + 6\sqrt 5 \left| y \right|\\ = \left\{ \begin{array}{l}6\sqrt 5 y + 6\sqrt 5 y\;\;\;\;khi\;\;\;y \ge 0\\6\sqrt 5 y - 6\sqrt 5 y\;\;\;\;khi\;\;y < 0\end{array} \right.\\ = \left\{ \begin{array}{l}12\sqrt 5 y\;\;\;khi\;\;y \ge 0\\0\;\;\;\;\;\;\;\;\;\;khi\;\;y < 0\end{array} \right..\end{array}\)

\(\begin{array}{l}d)\;3x\sqrt {72x}  - 9\sqrt {50{x^3}} \;\;\;\left( {x \ge 0} \right)\\ = 3x\sqrt {{6^2}.2x}  - 9\sqrt {{5^2}.2.{x^2}.x} \\ = 3x.6\sqrt {2x}  - 9.5.\left| x \right|\sqrt {2x} \\ = 18x\sqrt {2x}  - 45x\sqrt {2x} \\ =  - 27x\sqrt {2x} .\end{array}\)