Задание 7, стр 9

Question

Найдите значение выражения.
$$a)\sqrt{36}\cdot\sqrt{25};\quad b) 4 \cdot \sqrt{9}-3 \cdot \sqrt{4};\quad c) \sqrt{0,09}+\sqrt{0,25};\quad d) \sqrt{0,25-0,09}$$
$$e) \sqrt{\frac{49}{81}}: \sqrt{\frac{25}{9}}\quad f) \frac{5}{6} \cdot \sqrt{\frac{9}{25}}+\sqrt{5 \frac{1}{16}}\quad g) \sqrt{1,(7)}+\sqrt{0,(4)}\quad h) \sqrt{1,(8)+0,(8)}$$

Gimli · Accepted Answer

a. $$\sqrt{36} \cdot \sqrt{25}=\sqrt{6^{2}} \cdot \sqrt{5^{2}}=6 \cdot 5=30.$$
b. $$4 \cdot \sqrt{9}-3 \sqrt{4}=4 \cdot \sqrt{3^{2}}-3 \sqrt{2^{2}}=4 \cdot 3-3 \cdot 2=12-6=6.$$
c. $$\sqrt{0,09}+\sqrt{0,25}=\sqrt{0,3^{2}}+\sqrt{0,5^{2}}=0,3+0,5=0,8.$$
d. $$\sqrt{0,25-0,09}=\sqrt{0,16}=\sqrt{0,4^{2}}=0,4.$$
e. $$\sqrt{\frac{49}{81}}: \sqrt{\frac{25}{9}}=\sqrt{\frac{7^{2}}{9^{2}}}: \sqrt{\frac{5^{2}}{3^{2}}}=\sqrt{\left(\frac{7}{9}\right)^{2}}: \sqrt{\left(\frac{5}{3}\right)^{2}}=\frac{7}{9}: \frac{5}{3}=\frac{7}{9} \cdot \frac{3}{5}=\frac{7}{3 \cdot 5}=\frac{7}{15}.$$
f. $$\frac{5}{6} \cdot \sqrt{\frac{9}{25}}+\sqrt{5 \frac{1}{16}}=\frac{5}{6} \cdot \sqrt{\frac{3^{2}}{5^{2}}}+\sqrt{\frac{81}{16}}=\frac{5}{6} \cdot \sqrt{\left(\frac{31}{5}\right)^{2}}+\sqrt{\frac{9^{2}}{4^{2}}}=\frac{5}{6} \cdot \frac{3}{5}+\sqrt{\left(\frac{9}{4}\right)^{2}}=$$
$$=\frac{5}{6} \cdot \frac{3}{5}+\frac{9}{4}=\frac{1}{2}+\frac{9}{4}=\frac{2}{4}+\frac{9}{4}=\frac{11}{4} = 2\frac{3}{4}.$$
g. Так как $1.(7) = 1\frac{7}{9}$ и $0.(4)=\frac{4}{9},$ то
$$\sqrt{1.(7)}+\sqrt{0.(4)} = \sqrt{1\frac{7}{9}}+\sqrt{\frac{4}{9}} = \sqrt{\frac{16}{9}}+\sqrt{\frac{2^2}{3^2}} = \sqrt{\frac{4^2}{3^2}}+\sqrt{\left(\frac{2}{3}\right)^2} = \sqrt{\left(\frac{4}{3}\right)^2}+\frac{2}{3} = \frac{4}{3}+\frac{2}{3} = \frac{6}{3} = 2.$$
h. Так как $1.(8) = 1\frac{8}{9}$ и $0.(8)=\frac{8}{9},$ то
$$\sqrt{1.(8)+0.(8)} = \sqrt{1\frac{8}{9}+\frac{8}{9}} = \sqrt{1\frac{16}{9}}= \sqrt{\frac{25}{9}}= \sqrt{\left(\frac{5}{3}\right)^2} = \frac{5}{3}.$$

Задание 7, стр 9

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Задание 7, стр 9

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