552 Вправа 552Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 161

552. Порівняйте числа:

1) ${\displaystyle 5\sqrt{2}=}$${\displaystyle \sqrt{5^2\cdot 2}=}$${\displaystyle \sqrt{25\cdot 2}=}$${\displaystyle \sqrt{50}}$; ${\displaystyle 7=}$${\displaystyle \sqrt{7^2}=}$${\displaystyle \sqrt{49}}$;

${\displaystyle \sqrt{50}>\sqrt{49}}$, отже, ${\displaystyle 5\sqrt{2}>7}$;

2) ${\displaystyle 4\sqrt{5}=}$${\displaystyle \sqrt{4^2\cdot 5}=}$${\displaystyle \sqrt{16\cdot 5}=}$${\displaystyle \sqrt{80}}$; ${\displaystyle 9=}$${\displaystyle \sqrt{9^2}=}$${\displaystyle \sqrt{81}}$;

${\displaystyle \sqrt{80}<\sqrt{81}}$, отже, ${\displaystyle 4\sqrt{5}<9}$;

3) ${\displaystyle 4\sqrt{2}=}$${\displaystyle \sqrt{4^2\cdot 2}=}$${\displaystyle \sqrt{16\cdot 2}=}$${\displaystyle \sqrt{32}}$;

${\displaystyle \sqrt{32}=}$${\displaystyle \sqrt{32}}$, отже, ${\displaystyle 4\sqrt{2}=}$${\displaystyle \sqrt{32}}$;

4) ${\displaystyle 10\sqrt{11}=}$${\displaystyle \sqrt{{10}^2\cdot 11}=}$${\displaystyle \sqrt{100\cdot 11}=}$${\displaystyle \sqrt{1100}}$; ${\displaystyle 11\sqrt{10}=}$${\displaystyle \sqrt{{11}^2\cdot 10}=}$${\displaystyle \sqrt{121\cdot 10}=}$${\displaystyle \sqrt{1210}}$;

${\displaystyle \sqrt{1100}<\sqrt{1210}}$, отже, ${\displaystyle 10\sqrt{11}<11\sqrt{10}}$;

5) ${\displaystyle 6\sqrt{7}=}$${\displaystyle \sqrt{6^2\cdot 7}=}$${\displaystyle \sqrt{36\cdot 7}=}$${\displaystyle \sqrt{252}}$; ${\displaystyle 7\sqrt{6}=}$${\displaystyle \sqrt{7^2\cdot 6}=}$${\displaystyle \sqrt{49\cdot 6}=}$${\displaystyle \sqrt{294}}$;

${\displaystyle \sqrt{252}<\sqrt{294}}$, отже, ${\displaystyle 6\sqrt{7}<7\sqrt{6}}$;

6) ${\displaystyle 4\sqrt{0,25}=}$${\displaystyle \sqrt{4^2\cdot 0,25}=}$${\displaystyle \sqrt{16\cdot 0,25}=}$${\displaystyle \sqrt{4}}$;

${\displaystyle \sqrt{4}>\sqrt{0,49}}$, отже, ${\displaystyle 4\sqrt{0,25}>\sqrt{0,49}}$;

7) ${\displaystyle \frac{1}{4}=}$${\displaystyle 0,25}$;

${\displaystyle \sqrt{0,25}=}$${\displaystyle \sqrt{0,25}}$, отже, ${\displaystyle \sqrt{\frac{1}{4}}=}$${\displaystyle \sqrt{0,25}}$;

8) ${\displaystyle 2\frac{1}{4}=}$${\displaystyle 2,25<5}$;

${\displaystyle \sqrt{2,25}<\sqrt{5}}$, отже, ${\displaystyle \sqrt{2\frac{1}{4}}<\sqrt{5}}$;

9) ${\displaystyle 2\sqrt{3}=}$${\displaystyle \sqrt{2^2\cdot 3}=}$${\displaystyle \sqrt{4\cdot 3}=}$${\displaystyle \sqrt{12}}$;

${\displaystyle \sqrt{4,5}<\sqrt{12}}$, отже, ${\displaystyle \sqrt{4,5}<2\sqrt{3}}$.