Нужна помощь решить примеры какие сможете,буду благодарна 10 класс. Даю много пунктов.

Question 1

Question 2

1
1)
$\frac{15^{2/3}*3^{7/3}}{5^{-1/3}} = \frac{5^{2/3}*3^{2/3}*3^{7/3}}{5^{-1/3}} =\frac{5^{2/3}*3^{2/3+7/3}}{5^{-1/3}} =\frac{5^{2/3}*3^{9/3}}{5^{-1/3}} = \\ =\frac{5^{2/3}*3^{3}}{5^{-1/3}} =\frac{5^{2/3}*27}{5^{-1/3}} =5^{2/3-(-1/3)}*27=5^{2/3+1/3}*27=5^{3/3}*27= \\ =5*27=135$

2)(4/5)⁻²-(1/27)¹/³+4*379⁰=(5/4)²-1/∛27+4*1=25/16-1/3+4=75/48-16/48+4=
=59/48+4=1 11/48+4=5 11/48

3)(∛128+∛(1/4)):∛2=(∛128+∛(1/4))*1/∛2=∛128/∛2+∛(1/4)/∛2=∛64+∛1/8=4-1/2=3 1/2=3.5

2)
1)∛ab²/c * ∛a²b/c²=∛a³b³/c³=ab/c

2) $\frac{a^{-3}*a^{7/3}}{a^{1/3}} = a^{-3+7/3-1/3} =a^{-3+6/3} =a^{-3+2} =a^{-1} =1/a$

3) $\frac{a-9a^{1/2}}{7a^{1/4}+21} = \frac{a(1-9a^{-1/2})}{7(a^{1/4}+3)}$
что-то тут моя фантазия умерла

4)
$\sqrt[5]{(2/9)^3} i \sqrt[5]{(1/4)^3} \\ \sqrt[5]{(8/36)^3} i \sqrt[5]{(9/36)^3} \\ \sqrt[5]{(8/36)^3} < \sqrt[5]{(9/36)^3} \\$

5) (∛a+∛b)²-(∛a-∛b)²=(∛a+∛b+∛a-∛b)(∛a+∛b-(∛a-∛b))=
=(∛a+∛a)(∛b+∛b)=2∛a*2∛b=4∛ab

ndusha_zn · Answer 1 · 2018-03-21T05:04:18+0000

1
1)
$\frac{15^{2/3}*3^{7/3}}{5^{-1/3}} = \frac{5^{2/3}*3^{2/3}*3^{7/3}}{5^{-1/3}} =\frac{5^{2/3}*3^{2/3+7/3}}{5^{-1/3}} =\frac{5^{2/3}*3^{9/3}}{5^{-1/3}} = \\ =\frac{5^{2/3}*3^{3}}{5^{-1/3}} =\frac{5^{2/3}*27}{5^{-1/3}} =5^{2/3-(-1/3)}*27=5^{2/3+1/3}*27=5^{3/3}*27= \\ =5*27=135$

2)(4/5)⁻²-(1/27)¹/³+4*379⁰=(5/4)²-1/∛27+4*1=25/16-1/3+4=75/48-16/48+4=
=59/48+4=1 11/48+4=5 11/48

3)(∛128+∛(1/4)):∛2=(∛128+∛(1/4))*1/∛2=∛128/∛2+∛(1/4)/∛2=∛64+∛1/8=4-1/2=3 1/2=3.5

2)
1)∛ab²/c * ∛a²b/c²=∛a³b³/c³=ab/c

2) $\frac{a^{-3}*a^{7/3}}{a^{1/3}} = a^{-3+7/3-1/3} =a^{-3+6/3} =a^{-3+2} =a^{-1} =1/a$

3) $\frac{a-9a^{1/2}}{7a^{1/4}+21} = \frac{a(1-9a^{-1/2})}{7(a^{1/4}+3)}$
что-то тут моя фантазия умерла

4)
$\sqrt[5]{(2/9)^3} i \sqrt[5]{(1/4)^3} \\ \sqrt[5]{(8/36)^3} i \sqrt[5]{(9/36)^3} \\ \sqrt[5]{(8/36)^3} < \sqrt[5]{(9/36)^3} \\$

5) (∛a+∛b)²-(∛a-∛b)²=(∛a+∛b+∛a-∛b)(∛a+∛b-(∛a-∛b))=
=(∛a+∛a)(∛b+∛b)=2∛a*2∛b=4∛ab