Задание 12, стр 25

Question

Упростите.
$$	ext{a) $\sqrt{2 x}+\sqrt{9 y}+\sqrt{18 x}\quad \mathrm{b}) 2 \sqrt{27}+\sqrt{48 x}+\sqrt{12 x}$}$$
$$	ext{c) $\sqrt{24}+\sqrt{12}-\sqrt{6}+\sqrt{3}\quad \mathrm{d}) \sqrt{4}+\sqrt{50}-\sqrt{32}+\sqrt{36}$}$$

Gimli · Accepted Answer

$a)\quad\sqrt{2 x}+\sqrt{9 y}+\sqrt{18 x}=\sqrt{2} \cdot \sqrt{x}+\sqrt{9} \cdot \sqrt{y}+\sqrt{18} \cdot \sqrt{x}=\sqrt{2} \cdot \sqrt{x}+3 \sqrt{y}+\sqrt{9} \cdot \sqrt{2} \cdot \sqrt{x}=\sqrt{2} \cdot \sqrt{x}+$
$$+3 \sqrt{y}+3 \cdot \sqrt{2} \cdot \sqrt{x}=\sqrt{2} \cdot \sqrt{x}+3 \cdot \sqrt{2} \cdot \sqrt{x}+3 \sqrt{y}=\sqrt{2} \cdot \sqrt{x}(1+3)+3 \sqrt{y}=\sqrt{2 x} \cdot 4+3 \sqrt{y}=4 \sqrt{2 x}+3 \sqrt{y}.$$
$	ext{b) $\qquad 2 \sqrt{27}+\sqrt{48 x}+\sqrt{12 x}=2 \cdot \sqrt{9} \cdot \sqrt{3}+\sqrt{12 x}\cdot \sqrt{4}$
$+\sqrt{12 x}=2 \cdot 3 \cdot \sqrt{3}+\sqrt{12 x}(\sqrt{4}+1)=6 \sqrt{3}+$}$
$$+\sqrt{12} \cdot \sqrt{x} \cdot 3=6 \sqrt{3}+\sqrt{4} \cdot \sqrt{3} \cdot \sqrt{x} \cdot 3=6 \sqrt{3}+\sqrt{4} \cdot 3 \cdot \sqrt{3 x}=6 \sqrt{3}+6 \sqrt{3 x}.$$
$	ext{c) $\quad \sqrt{24}+\sqrt{12}-\sqrt{6}+\sqrt{3}=\sqrt{3} \cdot \sqrt{8}+\sqrt{3} \cdot \sqrt{4}\cdot\sqrt{3} \cdot \sqrt{2}+\sqrt{3}=\sqrt{3}(\sqrt{8}+\sqrt{4}-\sqrt{2}+1)=$}$
$$=\sqrt{3}(\sqrt{8}+2-\sqrt{2}+1)=\sqrt{3}(3+\sqrt{8}-\sqrt{2})=\sqrt{3}(3+\sqrt{2}(\sqrt{4}-1))=\sqrt{3}(3+\sqrt{2}(2-1))\sqrt{3}(3+\sqrt{2} \cdot 1)=$$
$$=\sqrt{3}(3+\sqrt{2})=3 \sqrt{3}+\sqrt{6}.$$
  $	ext{d) $\quad \sqrt{4}+\sqrt{50}-\sqrt{32}+\sqrt{36}=2+\sqrt{25} \cdot \sqrt{2}-\sqrt{16} \cdot \sqrt{2}$ $+6=2+5 \sqrt{2}-4 \sqrt{2}+6=8+\sqrt{2}$.}$

Задание 12, стр 25

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Задание 12, стр 25

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