$a\text{)}\left(x+2\right)\left(x-2\right)-2x>x^2+8$  

$x^2-4-2x>x^2+8\mid -x^2$  

$-4-2x>8\mid +4$  

$-2x>12\mid :\left(-2\right)$  

$x<-6$  

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$b\text{)}3x^2-3{\left(1-x\right)}^2-6<-3x$  

$3x^2-3\left(1-2x+x^2\right)-6<-3x$  

$3x^2-3+6x-3x^2-6<-3x$  

$6x-9<-3x\mid +3x$  

$9x-9<0\mid +9$  

$9x<9\mid :9$  

$x<1$  

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$c\text{)}{\left(x+3\right)}^2>x\left(x+6\right)$  

$x^2+6x+9>x^2+6x$  

$9>0$  

Każda liczba spełnia tę nierówność

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$d\text{)}{\left(y+\frac{1}{2}\right)}^2\ge y\left(y-\frac{3}{4}\right)$  

$y^2+y+\frac{1}{4}\ge y^2-\frac{3}{4}y\mid -y^2$  

$y+\frac{1}{4}\ge -\frac{3}{4}y\mid +\frac{3}{4}y$  

$1\frac{3}{4}y+\frac{1}{4}\ge 0$  

$\frac{7}{4}y+\frac{1}{4}\ge 0\mid -\frac{1}{4}$  

$\frac{7}{4}y>-\frac{1}{4}\mid \cdot 4$  

$7y>-1\mid :7$  

$y>-\frac{1}{7}$  

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$e\text{)}\frac{4y-3}{6}+3\le \frac{3y}{2}+\frac{5}{8}\mid \cdot 24$  

$4\left(4y-3\right)+72\le 12\cdot 3y+3\cdot 5$  

$16y-12+72\le 36y+15\mid -16y$  

$60\le 20y+15\mid -15$  

$45\le 20y\mid :20$  

$2,25\le y$  

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$f\text{)}\left(x-3\right)\left(x+3\right)>\frac{x+2}{2}+1+x^2$  

$x^2-9>\frac{x+2}{2}+1+x^2\mid -x^2$  

$-9>\frac{x+2}{2}+1\mid \cdot 2$  

$-18>x+2+2$  

$-18>x+4\mid -4$  

$-22>x$