Question

Помогите пожалуйста не могу сделать. 3.26, 3.27​

Answer 1

$1a)x>4\\\\(x-4)*\sqrt{x^{2}-8x+16}=(x-4)*\sqrt{(x-4)^{2}}=(x-4)*|x-4|=(x-4)*(x-4)=(x-4)^{2}\\\\1b)x<4\\\\(x-4)*\sqrt{x^{2}-8x+16}=(x-4)*\sqrt{(x-4)^{2}}=(x-4)*|x-4|=(x-4)(4-x)=-(x-4)^{2}$

$2a)x>-1\\\\(x+1)*\frac{1}{\sqrt{x^{2}+2x+1 }} =(x+1)*\frac{1}{\sqrt{(x+1)^{2}}}=(x+1)*\frac{1}{|x+1|}=(x+1)*\frac{1}{x+1}=1\\\\2b)x<-1\\\\(x+1)*\frac{1}{\sqrt{x^{2}+2x+1}}=(x+1)*\frac{1}{\sqrt{(x+1)^{2}}}=(x+1)*\frac{1}{|x+1|}=(x+1)*\frac{1}{-(x+1)}=-1$

3,27

$1)\sqrt{\sqrt{625}}=\sqrt{\sqrt{25^{2}}}=\sqrt{25}=5\\\\2)\sqrt{\sqrt{0,1296}}=\sqrt{\sqrt{0,36^{2}}}=\sqrt{0,36}=0,6\\\\3)6*(\sqrt{\frac{2}{3}})^{4}=6*[(\sqrt{\frac{2}{3}})^{2}]^{2}=6*(\frac{2}{3})^{2}=6*\frac{4}{9}=2\frac{2}{3}\\\\4)2*(-\sqrt{\frac{7}{12}})^{2}=2*(\frac{7}{12})=\frac{7}{6}=1\frac{1}{6}$