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)
1)=sin3x+3=sin4x
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x²-5x+4/x-1=x+1
Решаем как пропорция:
x²-5x+4=(x-1)(x+1)
x²-5x+4=x²-1
x²-5x+4-x²+1=0
-5x+5=0
-5x=-5
x=1-корней нет т.к знаменатель не равен 0