Задание 4, стр 8

Question

Найдите значение корня.
ewline
$$a) \sqrt{6 \frac{1}{4}};\quad b) \sqrt{1 \frac{7}{9}};\quad c) \sqrt{0,(4)};\quad d) \sqrt{2,(7)};\quad e) \sqrt{5 .(4)}.$$

Gimli · Accepted Answer

$$\sqrt{6 \frac{1}{4}}=\sqrt{\frac{25}{4}}=\sqrt{\frac{5^{2}}{2^{2}}}=\sqrt{\left(\frac{5}{2}\right)^{2}}=\frac{5}{2} = 2\frac{1}{2} = 2,5; $$
$$\sqrt{1 \frac{7}{9}}=\sqrt{\frac{16}{9}}=\sqrt{\frac{4^{2}}{3^{2}}}=\sqrt{\left(\frac{4}{3}\right)^{2}}=\frac{4}{3} = 1\frac{1}{3} = 1,(3); $$
$$\sqrt{0,(4)}=\sqrt{\frac{4}{9}}=\sqrt{\frac{2^{2}}{3^{2}}}=\sqrt{\left(\frac{2}{3}\right)^{2}}=\frac{2}{3} = 0,(6); $$
$$\sqrt{2,(7)}=\sqrt{2 \frac{7}{9}}=\sqrt{\frac{25}{9}}=\sqrt{\left(\frac{5}{3}\right)^{2}}=\frac{5}{3} = 1\frac{2}{3} = 1,(6); $$
$$\sqrt{5(4)}=\sqrt{5 \frac{4}{9}}=\sqrt{\frac{49}{9}}=\sqrt{\left(\frac{7}{3}\right)^{2}}=\frac{7}{3} = 2\frac{1}{3} = 2,(3) .$$

Задание 4, стр 8

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Задание 4, стр 8

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