Question

Упростите выражение:
а)$\sqrt{0,81a} -\sqrt{0,01a}+\sqrt{1,44a}$
б)$\sqrt{2,25c}-\sqrt{0,16c}-\sqrt{0,01c}$
в)$\sqrt{72x}+\sqrt{288x}-\sqrt{450x}$
г)$\sqrt{1,2y}-3\sqrt{4,8y}-\sqrt{10,8y}$

Answer 1

Объяснение:

а)$\sqrt{0,81a} -\sqrt{0,01a}+\sqrt{1,44a}$

$\sqrt{0,81a} -\sqrt{0,01a}+\sqrt{1,44a} = \\ = \sqrt{ {0.9}^{2} a} -\sqrt{ {0.1}^{2} a}+\sqrt{ {1.2}^{2} a} = \\ = 0.9 \sqrt{a} - 0.1 \sqrt{a} + 1.2 \sqrt{a} = \\ = (0.9 - 0.1 + 1.2) \cdot \sqrt{a} = 2 \sqrt{a}$

б)$\sqrt{2,25c}-\sqrt{0,16c}-\sqrt{0,01c}$

$\sqrt{2,25c}-\sqrt{0,16c}-\sqrt{0,01c} = \\ \sqrt{c} \cdot(\sqrt{2,25}-\sqrt{0,16}-\sqrt{0,01}) = \\ = \sqrt{c} \cdot(\sqrt{ {1.5}^{2} }-\sqrt{ {0.4}^{2} }-\sqrt{ {0.1}^{2} }) = \\ = \sqrt{c} \cdot(1.5 - 0.4 - 0.1) = \sqrt{c} \cdot{1} = \sqrt{c}$

в)$\sqrt{72x}+\sqrt{288x}-\sqrt{450x}$

$\sqrt{72x}+\sqrt{288x}-\sqrt{450x} = \\ = \sqrt{36 \cdot 2x}+\sqrt{144 \cdot 2x}-\sqrt{225 \cdot 2x} = \\ = \sqrt{2x} \cdot( \sqrt{36}+\sqrt{144}-\sqrt{225}) = \\ = \sqrt{2x} \cdot( \sqrt{ {6}^{2} }+\sqrt{ {12}^{2} }-\sqrt{ {15}^{2} }) = \\ = \sqrt{2x} \cdot(6 + 12 - 15) = \sqrt{2x} \cdot3 = \\ = 3 \sqrt{2x}$

г)$\sqrt{1,2y}-3\sqrt{4,8y}-\sqrt{10,8y}$

$\sqrt{1,2y}-3\sqrt{4,8y}-\sqrt{10,8y} = \\ \small{ = \sqrt{4\cdot0.3y}-3\sqrt{16\cdot0.3y}-\sqrt{36\cdot0.3y}} = \\ = \sqrt{0.3y} \cdot( \sqrt{4} - 3 \sqrt{16} - \sqrt{36} ) = \\ = \sqrt{0.3y} \cdot(2 - 3\cdot4 - 6) = \\ = \sqrt{0.3y} \cdot( - 16) = - 16 \sqrt{0.3y}$