Упр.42.28 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)
Вычислите значение выражения:
- $$3^{1,2}\cdot3^{-0,7}\cdot3^{1,5};$$
- $$11^{-\frac{4}{3}}\cdot11^{-\frac{3}{4}}\cdot11^{\frac{1}{12}};$$
- $$36^{0,7}\cdot6^{-0,4};$$
- $$\frac{27^{\frac{1}{2}}}{3^{\frac{1}{2}}};$$
- $$0,125^{-\frac{1}{3}}+0,81^{-\frac{1}{2}}-0,216^{-\frac{2}{3}};$$
- $$(0,027^{\frac{4}{3}})^{-0,25}+256^{0,75}-(\operatorname{tg}^2\frac{\pi}{3})^{-1};$$
- $$625^{0,25}-(\operatorname{ctg}^2\frac{\pi}{4})^{-\frac{3}{7}}+(100^{\frac{1}{7}})^{3,5};$$
- $$\left((8^{\frac{1}{2}})^{-\frac{2}{3}}+9^{-\frac{1}{4}}\right)\left((\sqrt{32})^{-\frac{2}{5}}-(3^{\frac{3}{2}})^{-\frac{1}{3}}\right);$$
- $$\left(9^{\frac{1}{4}}-\left(0,5\left((0,5)^{\frac{1}{3}}\right)\right)^{-0,75}\right)\left(81^{0,125}+\left(\cos^2\frac{\pi}{4}\right)^{-1}\right);$$
- $$\left(4^{0,25}+\left(\sin\frac{\pi}{6}\right)^{-1,5}\right)^{-\frac{4}{3}}\left(4^{\frac{1}{4}}-(2\sqrt{2})^{-\frac{4}{3}}\right).$$
1) $$3^{1,2}\cdot 3^{-0,7}\cdot 3^{1,5}=3^{1,2-0,7+1,5}=3^2=9.$$
Ответ: $$9$$
2) $$11^{-\frac43}\cdot 11^{-\frac34}\cdot 11^{\frac1{12}}=11^{-\frac43-\frac34+\frac1{12}}=11^{-\frac{16}{12}-\frac{9}{12}+\frac{1}{12}}=11^{-2}=\frac1{121}.$$
Ответ: $$\frac1{121}$$
3) $$36^{0,7}\cdot 6^{-0,4}=(6^2)^{0,7}\cdot 6^{-0,4}=6^{1,4}\cdot 6^{-0,4}=6.$$
Ответ: $$6$$
4) $$\frac{27^{\frac12}}{3^{\frac12}}=\left(\frac{27}{3}\right)^{\frac12}=9^{\frac12}=3.$$
Ответ: $$3$$
5) $$0,125^{-\frac13}+0,81^{-\frac12}-0,216^{-\frac23}=\left(\frac18\right)^{-\frac13}+\left(\frac{81}{100}\right)^{-\frac12}-\left(\frac{27}{125}\right)^{-\frac23}$$
$$=2+ \frac{10}{9}-\frac{25}{9}=2-\frac{15}{9}=2-\frac53=\frac13.$$
Ответ: $$\frac13$$
6) $$\left(0,027^{\frac43}\right)^{-0,25}+256^{0,75}-\left(\tg^2\frac{\pi}{3}\right)^{-1}$$
$$=\left(\left(\frac{3}{10}\right)^3\right)^{-\frac14}+256^{\frac34}-\left(3\right)^{-1}$$
$$=\left(\frac{3}{10}\right)^{-\frac34}+4^3-\frac13= \frac{10}{3}+64-\frac13=67.$$
Ответ: $$67$$
7) $$625^{0,25}-\left(\ctg^2\frac{\pi}{4}\right)^{-\frac37}+\left(\sqrt[7]{100}\right)^{3,5}$$
$$=(5^4)^{\frac14}-(1^2)^{-\frac37}+(100^{\frac17})^{\frac72}=5-1+100^{\frac12}=5-1+10=14.$$
Ответ: $$14$$
8) $$\left((8^{\frac12})^{-\frac23}+9^{-\frac14}\right)\left((\sqrt{32})^{-\frac25}-(3^{\frac32})^{-\frac13}\right)$$
$$=\left(8^{-\frac13}+9^{-\frac14}\right)\left(32^{-\frac15}-3^{-\frac12}\right)$$
$$=\left(2^{-1}+3^{-\frac12}\right)\left(2^{-1}-3^{-\frac12}\right)=2^{-2}-3^{-1}=\frac14-\frac13=-\frac1{12}.$$
Ответ: $$-\frac1{12}$$
9) $$\left(9^{\frac14}-\left(0,5\sqrt[3]{0,5}\right)^{-0,75}\right)\left(81^{0,125}+\left(\cos^2\frac{\pi}{4}\right)^{-1}\right)$$
$$=\left(3^{\frac12}-\left(0,5^{\frac43}\right)^{-\frac34}\right)\left(3^{\frac12}+\left(\frac12\right)^{-1}\right)$$
$$=\left(3^{\frac12}-0,5^{-1}\right)\left(3^{\frac12}+0,5^{-1}\right)=\left(\sqrt3-2\right)\left(\sqrt3+2\right)=3-4=-1.$$
Ответ: $$-1$$
10) $$\left(4^{0,25}+\left(\left(\sin\frac{\pi}{6}\right)^{-1,5}\right)^{-\frac43}\right)\left(4^{\frac14}-(2\sqrt2)^{-\frac43}\right)$$
$$=\left(2^{\frac12}+\left(\frac12\right)^2\right)\left(2^{\frac12}-\left(2^{\frac32}\right)^{-\frac43}\right)$$
$$=\left(\frac1{\sqrt2}+\frac14\right)\left(\frac1{\sqrt2}-\frac14\right)=\left(\frac1{\sqrt2}\right)^2-\left(\frac14\right)^2=\frac12-\frac1{16}=\frac7{16}.$$
Ответ: $$\frac7{16}$$
