(x+1)^{2}-5\\\\10x^{2}-10x>x^{2}+2x+1-5\\\\10x^{2}-10x-x^{2}-2x+4>0\\\\9x^{2}-12x+4>0\\\\9x^{2}-12x+4=0\\\\D=(-12)^{2}-4*9*4=144-144=0\\\\x=\frac{12}{18}=\frac{2}{3}\\\\9(x-\frac{2}{3})^{2} >0\\\\(x-\frac{2}{3})^{2}>0\\\\x\in(-\infty;\frac{2}{3})\cup(\frac{2}{3};+\infty)" alt="6)10x(x-1)>(x+1)^{2}-5\\\\10x^{2}-10x>x^{2}+2x+1-5\\\\10x^{2}-10x-x^{2}-2x+4>0\\\\9x^{2}-12x+4>0\\\\9x^{2}-12x+4=0\\\\D=(-12)^{2}-4*9*4=144-144=0\\\\x=\frac{12}{18}=\frac{2}{3}\\\\9(x-\frac{2}{3})^{2} >0\\\\(x-\frac{2}{3})^{2}>0\\\\x\in(-\infty;\frac{2}{3})\cup(\frac{2}{3};+\infty)" align="absmiddle" class="latex-formula">
0\\\\x^{2}+9x-10>0\\\\(x-1)(x+10)>0" alt="8)\frac{x-1}{4}-\frac{2x-3}{2}<\frac{x^{2}+3x }{8}\\\\2(x-1)-4(2x-3)<x^{2}+3x \\\\2x-2-8x+12<x^{2}+3x\\\\x^{2}+3x+6x-10>0\\\\x^{2}+9x-10>0\\\\(x-1)(x+10)>0" align="absmiddle" class="latex-formula">
+ - +
_______________₀_____________₀___________
- 10 1
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Ответ : x ∈ (- ∞ ; - 10) ∪ (1 ; + ∞)
10) (x² - 5x)² + 14(x² - 5x) + 40 ≥ 0
x² - 5x = m
m² + 14m + 40 ≥ 0
(m + 4)(m + 10) ≥ 0
+ - +
_________[- 10]___________[- 4]_____________
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1) m ≤ - 10
x² - 5x ≤ - 10
x² - 5x + 10 ≤ 0
x² - 5x + 10 = 0
D = (-5)² - 4 * 10 = 25 - 40 = - 15 < 0
решений нет
2) m ≥ - 4
x² - 5x ≥ - 4
x² - 5x + 4 ≥ 0
(x - 4)(x - 1) ≥ 0
+ - +
_________[1]_________[ 4]___________
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x ∈ (- ∞ ; 1] ∪ [4 ; + ∞)