Question

помогите решить пожалуйста с решением . все кроме первого ​

Answer 1

Пошаговое объяснение:

2) 20

3) 50

4) 80

5) 6

$\sqrt{6}$

7) 18

Answer 2

$\tt\displaystyle\frac{\sqrt[4]{16*81} \,*\sqrt{12} }{\sqrt{3} }=\frac{\sqrt[4]{4^{2} \,*9^{2} }*\sqrt{12}  }{\sqrt{3} } =\frac{\sqrt{4*9}\,*\sqrt{12}  }{\sqrt{3} } =\\ \\ \\ \frac{\sqrt{4*9*4*3} }{\sqrt{3} } =\sqrt{4*4*9} =\sqrt{4^{2}*3^{2}  } =4*3=12$

.........

$\tt\displaystyle\frac{\sqrt[3]{108}\,*\sqrt[6]{27*256}  }{\sqrt{12} } =\frac{\sqrt[3]{27*4}\,*\sqrt[6]{3^{3} *4^{3}*4 }  }{\sqrt{12} } =\frac{\sqrt[3]{3^{3} } \,*\sqrt[3]{4} \,*\sqrt{3*4} \,*\sqrt[3]{4} }{\sqrt{12} } =\\ \\ \\ \frac{3*\sqrt[3]{4*4} \,*\sqrt{12}  }{\sqrt{12} } =3\sqrt[3]{16} =3\sqrt[3]{8*2} =3\sqrt[3]{2^{3}*2 } =3*2\sqrt[3]{2} =6\sqrt[3]{2}$

..........

$\tt\displaystyle\frac{\sqrt[3]{72}\,*\sqrt{108}  }{\sqrt[6]{192} } =\frac{\sqrt[3]{9*8} \,*\sqrt{36*3}  }{\sqrt[6]{64*3} } =\frac{\sqrt[3]{3^{3}*2^{3}  }\,*6\sqrt{3}  }{\sqrt[6]{2^{6}*3 } } =\frac{3*2*6\sqrt{3} }{2\sqrt[6]{3} } =\frac{18\sqrt{3} }{\sqrt[6]{3} } =\frac{18\sqrt[6]{3^{3} } }{\sqrt[6]{3} } =18\sqrt[6]{9}=18\sqrt[6]{3^{2} } =18\sqrt[3]{3}$

...........

$\tt\displaystyle\frac{\sqrt{96} \,*\sqrt[3]{36} }{\sqrt[6]{6} } =\frac{\sqrt{16*6}\,*\sqrt[3]{6^{2} }  }{\sqrt[6]{6} } =\frac{2\sqrt{6} \,*\sqrt[6]{6^{4} } }{\sqrt[6]{6} } =2\sqrt[6]{6^{3} } \,*\sqrt[6]{6^{5} }=2\sqrt[6]{6^{8} }=\\   \\ \\ 2\sqrt[3]{6^{4} } =2\sqrt[3]{6^{3} *6} =2*6\sqrt[3]{6} =12\sqrt[3]{6}$