Задание 6, стр 9

Question

Вычислите.
$$a) \sqrt{144}+\sqrt{25}\quad b) \sqrt{144+25};\quad c) \sqrt{11+\sqrt{25}}; \quad d) \sqrt{29-\sqrt{16}}$$
$$ e) \sqrt{289}-\sqrt{64};\quad f) \sqrt{289-64};\quad g) \sqrt{49-\sqrt{169}};\quad h) \sqrt{\sqrt{144}+\sqrt{16}}$$

Gimli · Accepted Answer

$$
\begin{aligned}
&	ext { a. } \sqrt{144}+\sqrt{25}=\sqrt{12^{2}}+\sqrt{5^{2}}=12+5=17 . \
&	ext { b. } \sqrt{144+25}=\sqrt{169}=\sqrt{13^{2}}=13 \
&	ext { c. } \sqrt{11+\sqrt{25}} = \sqrt{11+\sqrt{5^2}} = \sqrt{11+5} = \sqrt{16}= \sqrt{4^2} = 4\
&	ext { d. } \sqrt{29-\sqrt{16}}=\sqrt{29-\sqrt{4^{2}}}=\sqrt{29-4}=\sqrt{25}=\sqrt{5^{2}}=5 \
&	ext { e. } \sqrt{289}-\sqrt{64}=\sqrt{17^{2}}-\sqrt{8^{2}}=17-8=9 . \
&	ext { f. } \sqrt{289-64}=\sqrt{225}=\sqrt{15^{2}}=15 . \
&	ext { g. } \sqrt{49-\sqrt{169}}=\sqrt{49-13}=\sqrt{36}=\sqrt{6^{2}}=6 \
&	ext { h. } \sqrt{\sqrt{144}+\sqrt{16}}=\sqrt{\sqrt{12^{2}}+\sqrt{4^{2}}}=\sqrt{12+4}=\sqrt{16}=\sqrt{4^{2}}=4
\end{aligned}
$$

Задание 6, стр 9

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Задание 6, стр 9

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