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Решите задачу:

$1)\; \; \frac{7x^2+28x+21}{2x^3+9x^2+10x+3}=\frac{7\cdot (x+3)(x+1)}{2(x+1)(x+3)(x+0,5)}=\frac{7}{2(x+0,5)}=\frac{7}{2x+1}\; ;\\\\\star \; \; 7x^2+28x+21=7\cdot (x^2+4x+3)\; ,\\\\x^2+4x+3=0\; \; \to \; \; x_1=-3\; ,\; x_2=-1\; \; (teorema\; Vieta)\; ,\\\\7\cdot (x^2+4x+3)=7\cdot (x+3)(x+1)\; .\\\\\star \; x=-1:\; 2x^3+9x^2+10x+3=-2+9-10+3=0\; \to \; x=-1\; \; koren'\; ,\\\\2x^3+9x^2+10x+3=(x+1)(\underbrace {2x^2+7x+3}_{x_1=-3,x_2=-0,5})=2(x+1)(x+3)(x+0,5)\; .$

$2)\; \; \frac{n^4-7n^3+4n^2+7n-5}{n^4-8n^3+12n^2-5n}=\frac{(n-1)(n+1)(n^2-7n+5)}{n(n-1)(n^2-7n+5)}=\frac{n+1}{n}\; ;\\\\\star \; \; n=1:\; n^4-7n^2+4n^2+7n-5=1-7+4+7-5=0\; ,\; n=1\; koren'\; ,\\\\n^4-7n^3+4n^2+7n-5=(n-1)(n^3-6n^2-2n+5)\; ,\\\\\star \; n=-1:\; n^3-6n^2-2n+5=-1-6=2+5=0\; \to \; n=-1\; koren'\; ,\\\\ n^3-6n^2-7n+5=(n+1)(n^2-7n+5)\; ,\\\\\star \; \; n^4-8n^3+12n^2-5n=n(n^3-8n^2+12n-5)\; ,\\\\\star \; n=1:\; n^3-8n^2+12n-5=1-8+12-5=0\; \to \; n=1\; koren'\; ,\\\\n^3-8n^2+12n-5=(n-1)(n^2-7n+5)$