351 Вправа 351Алгебра 8 клас Бевз 2025 · § 7 · стор. 70

351. Спрости вираз (1 − a²)/((1 + ax)² − (a + x)²) · (x + x²)/(1 − x) − 1/(1 − x)².

1) ${\displaystyle {(1+ax)}^2-{(a+x)}^2}$=

=${\displaystyle (1+ax-a-x)(1+ax+a+x)}$=

=${\displaystyle ((1-a)-x(1-a))((1+a)+x(1+a))}$=

=${\displaystyle (1-a)(1-x)(1+a)(1+x)}$;

2) ${\displaystyle \frac{1-a^2}{(1-a)(1-x)(1+a)(1+x)}\cdot \frac{x+x^2}{1-x}}$=

=${\displaystyle \frac{(1-a^2)\cdot (x+x^2)}{(1-a)(1-x)(1+a)(1+x)\cdot (1-x)}}$=

=${\displaystyle \frac{x\cancel{\left(a+1\right)}\cancel{\left(1-a\right)}\cancel{\left(x+1\right)}}{\cancel{\left(a+1\right)}\cancel{\left(1-a\right)}\cancel{\left(x+1\right)}{(1-x)}^2}=}$${\displaystyle \frac{x}{{(x-1)}^2}}$;

3) ${\displaystyle \frac{x}{{(x-1)}^2}-\frac{1}{{(1-x)}^2}=}$${\displaystyle \frac{x}{{(x-1)}^2}-\frac{1}{{(x-1)}^2}=}$${\displaystyle \frac{x-1}{{(x-1)}^2}}$=

=${\displaystyle \frac{\cancel{\left(x-1\right)}}{(x-1)\cancel{\left(x-1\right)}}=}$${\displaystyle \frac{1}{x-1}}$.