332 Вправа 332Алгебра 8 клас Бевз 2025 · § 7 · стор. 68

332. Спрости вираз.

а) ${\displaystyle (\frac{2x-1}{x+1}-x)\cdot (\frac{3}{x^3+1}-\frac{1}{x+1}-\frac{3}{x^2-x+1})}$;

1) ${\displaystyle \frac{2x-1}{x+1}-\frac{x^{\mathrm{\setminus }x+1}}{1}=}$${\displaystyle \frac{2x-1-x(x+1)}{x+1}=}$${\displaystyle \frac{2x-1-x^2-x}{x+1}}$=

=${\displaystyle \frac{x-x^2-1}{x+1}=}$${\displaystyle -\frac{x^2-x+1}{x+1}}$;

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

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

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

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

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

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

б) ${\displaystyle (1+\frac{b^2+c^2-a^2}{2bc}):\frac{a+b+c}{2abc}+a^2}$;

1) ${\displaystyle 1+\frac{b^2+c^2-a^2}{2bc}=}$${\displaystyle \frac{2bc+b^2+c^2-a^2}{2bc}}$=

=${\displaystyle \frac{{(b+c)}^2-a^2}{2bc}=}$${\displaystyle \frac{(b+c-a)(b+c+a)}{2bc}}$;

2) ${\displaystyle \frac{(b+c-a)(b+c+a)}{2bc}:\frac{a+b+c}{2abc}}$=

=${\displaystyle \frac{(b+c-a)(b+c+a)}{2bc}\cdot \frac{2abc}{a+b+c}}$=

=${\displaystyle \frac{(b+c-a)\cancel{\left(a+b+c\right)}\cdot a\cdot \cancel{2bc}}{\cancel{2bc}\cancel{\left(a+b+c\right)}}=}$${\displaystyle a(b+c-a)}$;

3) ${\displaystyle a(b+c-a)+a^2=}$${\displaystyle ab+ac-a^2+a^2}$=

=${\displaystyle ab+ac=}$${\displaystyle a\left(b+c\right)}$.