1 1*√3 √3
```` = ````````` = ````````
√3 √3 *√3 3
√10 + 8 (√10 +8)(√10 +8) (√10+8)² 10+16√10 +64 -37-8√10
``````````` = ````````````````````````` = `````````` = ```````````````````` = ``````````
√10 - 8 (√10 -8)(√10+8) 10 - 64 - 54 27
1 5+√5 5+√5 5+√5
````` = ``````````````````` = ``````````` = ````````````
5-√5 (5- √5)(5+√5) 25 -5 20
1) (2^2)^{2/3} *2^2*(2^2)^{-2}=
2^{4/3}*2^2*2^{-4}=2^{4/3+2-4}=2^{-2/3}=
\frac{1}{2^{2/3}}= \frac{1}{\sqrt[3]{2^2}}= \frac{1}{\sqrt[3]{4}}
2)2^ {-4} *2^3*(2^2)^3*2^{1/2}=2^{11/2}= \sqrt[11]{2^2}=\sqrt[11]{4}
3)(2^3)^3 *2^{-1/4}*2=2^{9-1/4+1}=2^{39/4}= \sqrt[4]{2^{39} }
4)
![10^2*(10^2)^{-2}*10=10^{2-4+1}=2^{-1}= \frac{1}{10}](https://tex.z-dn.net/?f=10%5E2%2A%2810%5E2%29%5E%7B-2%7D%2A10%3D10%5E%7B2-4%2B1%7D%3D2%5E%7B-1%7D%3D%20%5Cfrac%7B1%7D%7B10%7D%20)
5)
![a^{1/4-2+4+1/2} =a^{11/4}= \sqrt[4]{a^{11}}](https://tex.z-dn.net/?f=a%5E%7B1%2F4-2%2B4%2B1%2F2%7D%20%3Da%5E%7B11%2F4%7D%3D%20%5Csqrt%5B4%5D%7Ba%5E%7B11%7D%7D%20)
6)
![b^{2/3}: b : b^{1/3}*4^4= b^{2/3-1-1/3+4}=b^{10/3}= \sqrt[3]{b^{10}}](https://tex.z-dn.net/?f=b%5E%7B2%2F3%7D%3A%20b%20%3A%20b%5E%7B1%2F3%7D%2A4%5E4%3D%20b%5E%7B2%2F3-1-1%2F3%2B4%7D%3Db%5E%7B10%2F3%7D%3D%20%5Csqrt%5B3%5D%7Bb%5E%7B10%7D%7D%20)
Ответ:
Объяснение:
1) преобразуем у=8x^(-3)+3/x-4x^(3/2)+2x^7, y'=-24x^(-4)-3/x^2-6x^(1/2)+14x^6=-24/x^4-3/x^2-6Vx+14x^6 (V-корень)
2)y'=1/2cos x/2
3) y'=-4(5x-3)' / (5x-3)^2=-20 /(5x-3)^2