Question

1) 28a - 32
$\leqslant 7a {}^{2} - 4$
сколе буде
​

Answer 1

$\boldsymbol {(x-y)^2=x^2-2xy +y^2}$

$\large \boldsymbol {} 28a -32\geq 7a^2-4 \\\\ 7a^2-28a+32-4\geq 0 \\\\ 7a^2-28a+28\geq 0 \ \ |\div 7 \\\\ a^2-4a+4\geq 0 \\\\ \underbrace{a^2}_{x^2} -2\underbrace{ \cdot 2 \cdot a}_{xy} +\underbrace{2^2}_{y^2}\geq 0 \\\\ (a-2)^2\geq 0 \\\\ znaki : \ \ \ \ <+++[2]+++> \\\\ x\in ( -\infty \ \ ; \ \ 2) \cup (2\ \ ; \ \ \infty ) => (-\infty \ \ ; \ \ \infty )=> \mathbb R \\\\ Otvet : x \in \mathbb R$