9 bài tập Rút gọn biểu thức (có lời giải)

a) Ta có $2\sqrt {3x} - 4\sqrt {3x} + 27 - 3\sqrt {3x} = (2 - 4 - 3)\sqrt {3x} + 27 = - 5\sqrt 3 x + 27$.

b) Ta có $3\sqrt {2x} - 5\sqrt {8x} + 7\sqrt {18x} + 28 = 3\sqrt {2x} - 10\sqrt {2x} + 21\sqrt {2x} + 28$

$= (3 - 10 + 21)\sqrt {2x} + 28 = 14\sqrt {2x} + 28$.

a) Ta có $\frac{2}{{{x^2} - {y^2}}}\sqrt {\frac{{3{{(x + y)}^2}}}{2}} = \frac{{|x + y|}}{{{x^2} - {y^2}}}\sqrt {\frac{{4.3}}{2}} = \frac{{x + y}}{{(x - y)(x + y)}}.\sqrt 6 = \frac{{\sqrt 6 }}{{x - y}}$ (vì $x + y > 0$) .

b) $\frac{2}{{2a - 1}}\sqrt {5{a^2}\left( {1 - 4a + 4{a^2}} \right)} = \frac{2}{{2a - 1}}\sqrt {5{a^2}{{(1 - 2a)}^2}}$

$= \frac{2}{{2a - 1}}|a|.|1 - 2a|\sqrt 5 = \frac{{2\sqrt 5 }}{{2a - 1}}.a.(2a - 1)\quad \left( {{\rm{ do }}a > \frac{1}{2}} \right) = 2\sqrt 5 {\rm{a}}.$

a). $\sqrt {18{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} = \sqrt {2.9{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} = 3\,\left| {\sqrt 2 - \sqrt 3 } \right|\sqrt 2 = 3\left( {\sqrt 3 - \sqrt 2 } \right)\sqrt 2 = 3\sqrt 6 - 6.$

b). $ab\sqrt {1 + \frac{1}{{{a^2}{b^2}}}} = ab\sqrt {\frac{{{a^2}{b^2} + 1}}{{{a^2}{b^2}}}} = \frac{{ab}}{{\,\left| {ab} \right|}}\sqrt {{a^2}{b^2} + 1}$$\, = \left\{ \begin{array}{l}\sqrt {{a^2}{b^2} + 1} \,\,\,neu\,ab > 0\,\,\,\\ - \sqrt {{a^2}{b^2} + 1} \,\,\,neu\,\,ab < 0.\end{array} \right.$

c). $\sqrt {\frac{a}{{{b^3}}} + \frac{a}{{{b^4}}}} = \sqrt {\frac{{ab + a}}{{{b^4}}}} = \frac{{\sqrt {ab + a} }}{{{b^2}}}.$

d). $\frac{{a + \sqrt {ab} }}{{\sqrt a + \sqrt b }} = \frac{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a + \sqrt b }} = \sqrt a .$

Cách khác: $\frac{{a + \sqrt {ab} }}{{\sqrt a + \sqrt b }} = \frac{{\left( {a + \sqrt {ab} } \right)\left( {\sqrt a - \sqrt b } \right)}}{{a - b}} = \frac{{\sqrt a \left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}{{a - b}} = \frac{{\sqrt a \left( {a - b} \right)}}{{a - b}} = \sqrt a .$

Ta có: $\frac{{2 + \sqrt 2 }}{{1 + \sqrt 2 }} = \frac{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}}{{1 + \sqrt 2 }} = \sqrt 2 ;$ $\frac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }} = \frac{{\sqrt 5 \left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }} = - \sqrt 5 ;$

$\frac{{2\sqrt 3 - \sqrt 6 }}{{\sqrt 8 - 2}} = \frac{{\sqrt 6 \left( {\sqrt 2 - 1} \right)}}{{2\left( {\sqrt 2 - 1} \right)}} = \frac{{\sqrt 6 }}{2};$ $\frac{{a - \sqrt a }}{{1 - \sqrt a }} = \frac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{1 - ra}} = - \sqrt a ;$

$$\frac{{p - 2\sqrt p }}{{\sqrt p - 2}} = \frac{{\sqrt p \left( {\sqrt p - 2} \right)}}{{\sqrt p - 2}} = \sqrt p .$$

a) $5\sqrt a - 4b\sqrt {25{a^3}} + 5\sqrt {16a{b^2}} - 2\sqrt {9a} = 5\sqrt a - 20ab\sqrt a + 20ab\sqrt a - 6\sqrt a = - \sqrt a .$

b) $5a\sqrt {64a{b^3}} - \sqrt 3 \sqrt {12{a^3}{b^3}} + 2ab\sqrt {9ab} - 5b\sqrt {81{a^3}b}$

$$= 40ab\sqrt {ab} - 6ab\sqrt {ab} + 6ab\sqrt {ab} - 45ab\sqrt {ab} = - 5ab\sqrt {ab} .$$