Question

Помогите пожалуйста дам 35 баллов

Answer 1

$1)\ \ 4,1<\sqrt{17}<4,2\\\\\\a)\ \ 4,1-4<\sqrt{17}-4<4,2-4\ \ ,\ \ \ 0,1<\sqrt{17}-4<0,2\\\\\\ b)\ \ \sqrt{68}=\sqrt{4\cdot 17}=2\sqrt{17}\ \ \ \to \ \ \ \ 8,2<2\sqrt{17}<8,4\\\\\\c)\ \ -\sqrt{153}=-\sqrt{9\cdot17}=-3\sqrt{17}\ \ \ \to \ \ \ 12,3<3\sqrt{17}<12,6\ \ \to \\\\-12,6<-3\sqrt{17} <-12,3$

$2a)\ \ 144a^2-120a+37>0\ \ ,\\\\(\ (12a)^2-2\cdot 12a\cdot 5+5^2\ )+12=\underbrace {(12a-5)^2}_{\geq 0}+\underbrace {12}_{>0}>0\ \ \ verno\\\\\\b)\ \ 121x^2-132x+39>0\\\\(\ (11x)^2-2\cdot 11x\cdot 6+6^2\ )+3=\underbrace{(11x-6)^2}_{\geq 0}+\underbrace {3}_{>0}>0\ \ \ verno$

$c)\ \ 225c^2>30c-6\ \ ,\ \ 225c^2-30c+6>0\ \ ,\\\\\\(\ (15c)^2-2\cdot 15c\cdot 1+1^2\ )+5=\underbrace {(15c-1)^2}_{\geq 0}+\underbrace {5}_{>0}>0\ \ \ verno\\\\\\d)\ \ 49a^2-6a>120a-91\ \ ,\ \ \ 49a^2-126a+91>0\ \ ,\\\\(\ (7a)^2-2\cdot 17a\cdot 9+9^2\ )+10=\underbrace{(7a-9)^2}_{\geq 0}+\underbrace {10}_{>0}>0\ \ \ \ verno$

$3)\ \ p>x\ ,\ \ p<l\ ,\ \ y>l\ \ \ \Rightarrow \ \ \ \ y>l>p>x\ \ \ \Rightarrow \\\\\\\dfrac{1}{x}>\dfrac{1}{p}>\dfrac{1}{l}>\dfrac{1}{y}$