Упр.42.28 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) 3^(1,2)·3^(-0,7)·3^(1,5);
2) 11^(-4/3)·11^(-3/4)·11^(1/12);
3) 36^(0,7)·6^(-0,4);
4) (27^(1/2))/3^(1/2);
5) 0,125^(-1/3)+0,81^(-1/2)-0,216^(-2/3);
6) (0,027^(4/3))^(-0,25)+256^(0,75)-(tg^2(?/3))^(-1);
7) 625^(0,25)-(ctg^2(?/4))^(-3/7)+(100^(1/7))^3,5;
8) ((8^(1/2))^(-2/3)+9^(-1/4))((v32)^(-2/5)-(3^(3/2))^(-1/3));
9) (9^(1/4)-(0,5((0,5)^(1/3)))^(-0,75))(81^(0,125)+(cos^2(?/4))^(-1));
10) (4^0,25+((sin(?/6))^(-1,5))^(-4/3))(4^(1/4)-(2v2)^(-4/3)).

1. $$3^{1,2}\cdot 3^{-0,7}\cdot 3^{1,5}=3^{1,2-0,7+1,5}=3^2=9.$$

2. $$11^{-\frac43}\cdot 11^{-\frac34}\cdot 11^{\frac1{12}}=11^{-\frac43-\frac34+\frac1{12}}=11^{-\frac{16}{12}-\frac{9}{12}+\frac1{12}}=11^{-2}=\frac1{121}.$$

3. $$36^{0,7}\cdot 6^{-0,4}=(6^2)^{0,7}\cdot 6^{-0,4}=6^{1,4}\cdot 6^{-0,4}=6.$$

4. $$\frac{27^{\frac12}}{3^{\frac12}}=\left(\frac{27}{3}\right)^{\frac12}=9^{\frac12}=3.$$

5. $$0,125^{-\frac13}+0,81^{-\frac12}-0,216^{-\frac23}=\left(\frac18\right)^{-\frac13}+\left(\frac{81}{100}\right)^{-\frac12}-\left(\frac{216}{1000}\right)^{-\frac23}$$
   $$=2+\frac{10}{9}-\frac{25}{9}=\frac13.$$

6. $$\left(0,027^{\frac43}\right)^{-0,25}+256^{0,75}-\left(\tg^2\frac{\pi}{3}\right)^{-1}$$
   $$=\left(\left(\frac{27}{1000}\right)^{\frac13}\right)^{-1}+256^{\frac34}-\left(3\right)^{-1}$$
   $$=\left(\frac{3}{10}\right)^{-1}+4^3-\frac13=\frac{10}{3}+64-\frac13=67.$$

7. $$625^{0,25}-\left(\ctg^2\frac{\pi}{4}\right)^{-\frac37}+\left(\sqrt[7]{100}\right)^{3,5}$$
   $$=(5^4)^{\frac14}-(1^2)^{-\frac37}+(100^{\frac17})^{\frac72}$$
   $$=5-1+100^{\frac12}=5-1+10=14.$$

8. $$\left((8^{\frac12})^{-\frac23}+9^{-\frac14}\right)\left((\sqrt{32})^{-\frac25}-(3^{\frac32})^{-\frac13}\right)$$
   $$=\left(8^{-\frac13}+9^{-\frac14}\right)\left((2^5)^{-\frac15}-3^{-\frac12}\right)$$
   $$=\left(2^{-1}+3^{-\frac12}\right)\left(2^{-1}-3^{-\frac12}\right)=2^{-2}-3^{-1}=\frac14-\frac13=-\frac1{12}.$$

9. $$\left(9^{\frac14}-\left(0,5\sqrt[3]{0,5}\right)^{-0,75}\right)\left(81^{0,125}+\left(\cos^2\frac{\pi}{4}\right)^{-1}\right)$$
   $$=\left(3^{\frac12}-\left(0,5^{\frac43}\right)^{-\frac34}\right)\left(3^{\frac12}+\left(\frac12\right)^{-1}\right)$$
   $$=\left(3^{\frac12}-0,5^{-1}\right)\left(3^{\frac12}+0,5^{-1}\right)=3-0,5^{-2}=3-4=-1.$$

10. $$\left(4^{0,25}+\left(\left(\sin\frac{\pi}{6}\right)^{-1,5}\right)^{-\frac43}\right)\left(4^{\frac14}-(2\sqrt2)^{-\frac43}\right)$$
    $$=\left(2^{\frac12}+\left(\frac12\right)^2\right)\left(2^{\frac12}-\left(2^{\frac32}\right)^{-\frac43}\right)$$
    $$=\left(2^{\frac12}+2^{-2}\right)\left(2^{\frac12}-2^{-2}\right)=2-2^{-4}=2-\frac1{16}=\frac{31}{16}.$$

Ответ

1) $$9$$; 2) $$\frac1{121}$$; 3) $$6$$; 4) $$3$$; 5) $$\frac13$$; 6) $$67$$; 7) $$14$$; 8) $$-\frac1{12}$$; 9) $$-1$$; 10) $$\frac{31}{16}$$.