Question

розвяжіть нерівність 4) 5x (x + 4) - (2x - 3)(2x + 3) > 30;
5) (3x - 7)(x + 2) - (x - 4)(x + 5)> 30;​

Answer 1

$5x(x+4)-(2x-3)(2x+3)>30\\\\\\5x^2+20x-(4x^2-9)>30\\\\\\5x^2+20x-4x^2+9>30\\\\\\x^2+20x+9>30\\\\\\x^2+20x+9-30>0\\\\\\x^2+20x-21>0\\\\\\x^2+20x-21=0\\\\\\D=20^2-4\cdot(-21)=400+84=484=22^2\\\\\\x_1=\dfrac{-20-22}{2}=\dfrac{-42}{2}=-21\\\\\\x_2=\dfrac{-20+22}{2}=\dfrac{2}{2}=1\\\\\\+++(-21)---(1)+++>\\\\\\\boxed{x\in(\ -\infty\ ;\ -21\ )\ \cup\ (\ 1\ ;\ +\infty\ )}$

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$(3x - 7)(x + 2) - (x - 4)(x + 5)>30\\\\\\(3x^2+6x-7x-14)-(x^2+5x-4x-20)>30\\\\\\3x^2+6x-7x-14-x^2-5x+4x+20>30\\\\\\2x^2-2x+6-30>0\\\\\\2x^2-2x-24>0\ \ \mid\div2\\\\\\x^2-x-12=0\\\\\\D=1+4\cdot12=1+48=49=7^2\\\\\\x_1=\dfrac{1-7}{2}=-\dfrac{6}{2}=-3\\\\\\x_2=\dfrac{1+7}{2}=\dfrac{8}{2}=4\\\\\\+++(-3)---(4)+++>\\\\\\\boxed{x\in(-\infty;-3)\cup(4;+\infty)}$