Question

Решите задание пожалуйста

Answer 1

Ответ:

$1)\ \ 4<5<9\ \ \ \Rightarrow \ \ \sqrt4<\sqrt5<\sqrt9\ \ \ \Rightarrow \ \ \ 2<\sqrt5<3\\\\4,84<5<5,29\ \ \Rightarrow \ \ \sqrt{4,84}<\sqrt5<\sqrt{5,29}\ \ \ \Rightarrow \ \ \ 2,2<\sqrt5<2,3\\\\(\, 2,2\ ;\ 2,3\ )\subset (\, 1,9\ ;\ 2,4)\ \ \ \Rightarrow \ \ \underline{\ \sqrt5\in (\, 1,9\ ;\ 2,4\, )\ }\\\\Otvet:\ A)\ .$

$2)\ \ \sqrt{23}\ \ ,\ \ 3\sqrt7=\sqrt{9\cdot 7}=\sqrt{63}\ \ ,\ \ 5\sqrt3=\sqrt{25\cdot 3}=\sqrt{75}\ \ ,\\\\23<63<75\ \ \ \Rightarrow \ \ \ \ \sqrt{23}<\sqrt{63}<\sqrt{75}\ \ \ \Rightarrow \ \ \ \underline{\ \sqrt{23}<3\sqrt7<5\sqrt{3}\ }$  

$3)\ \ \dfrac{\sqrt{363}}{\sqrt3}+\sqrt{(-5)^2}-\sqrt{8}\cdot \sqrt2=\sqrt{\dfrac{363}{3}}+|-5|-\sqrt{8\cdot 2}=\sqrt{121}+5-\sqrt{16}=\\\\\\=11+5-4=12$

$4a)\ \ 4x^2\sqrt{5x}=\Big[\ \sqrt{5x}\ \ \to \ \ x\geq 0\ \Big]=\sqrt{(4x^2)^2\cdot 5x}=\sqrt{16x^4\cdot 5x}=\sqrt{80x^5}\\\\\\b)\ \ \sqrt{98a^9}=\Big[\ a\geq 0\ \Big]=\sqrt{49\cdot 2\cdot a^8\cdot a}=\sqrt{7^2\cdot 2\cdot (a^4)^2\cdot a}=\\\\=7\cdot |\, a^4\, |\cdot \sqrt{2a}=7\, a^4\cdot \sqrt{2a}$

$5)\ \ \dfrac{a}{\sqrt7}=\dfrac{a\cdot \sqrt7}{\sqrt7\cdot \sqrt7}=\dfrac{a\, \sqrt7}{7}\\\\\\\dfrac{3}{\sqrt5+a}=\dfrac{3(\sqrt5-a)}{(\sqrt5+a)(\sqrt5-a)}=\dfrac{3\, (\sqrt5-a)}{5-a^2}$