313 Вправа 313Алгебра 8 клас Бевз 2025 · § 7 · стор. 66

313. Доведи тотожність.

а) ${\displaystyle \frac{{(2a-1)}^2}{{(2a+1)}^2-8a}=}$${\displaystyle 1}$;

знаменник: ${\displaystyle {(2a+1)}^2-8a}$=

=${\displaystyle 4a^2+4a+1-8a}$=

=${\displaystyle 4a^2-4a+1=}$${\displaystyle {(2a-1)}^2}$;

${\displaystyle \frac{{(2a-1)}^2}{{(2a-1)}^2}=}$${\displaystyle 1}$;

тотожність доведено;

б) ${\displaystyle \frac{1}{5x}-\frac{1}{x+y}\cdot (\frac{x+y}{5x}-x-y)=}$${\displaystyle 1}$;

1) ${\displaystyle \frac{x+y}{5x}-\frac{x^{\mathrm{\setminus }5x}}{1}-\frac{y^{\mathrm{\setminus }5x}}{1}}$=

=${\displaystyle \frac{x+y-5x^2-5xy}{5x}=}$${\displaystyle \frac{(x+y)(1-5x)}{5x}}$;

2) ${\displaystyle \frac{1}{x+y}\cdot \frac{(x+y)(1-5x)}{5x}=}$${\displaystyle \frac{(1-5x)\cancel{\left(x+y\right)}}{5x\cancel{\left(x+y\right)}}=}$${\displaystyle \frac{1-5x}{5x}}$;

3) ${\displaystyle \frac{1}{5x}-\frac{1-5x}{5x}=}$${\displaystyle \frac{1-(1-5x)}{5x}=}$${\displaystyle \frac{1-1+5x}{5x}=}$${\displaystyle \frac{\cancel{5x}}{\cancel{5x}}=}$${\displaystyle 1}$;

тотожність доведено;

в) ${\displaystyle \frac{8a^2+2}{{(2a+1)}^2+{(2a-1)}^2}=}$${\displaystyle 1}$;

знаменник: ${\displaystyle {(2a+1)}^2+{(2a-1)}^2}$=

=${\displaystyle 4a^2+4a+1+4a^2-4a+1=}$${\displaystyle 8a^2+2}$;

${\displaystyle \frac{8a^2+2}{8a^2+2}=}$${\displaystyle 1}$;

тотожність доведено;

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

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

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

=${\displaystyle \frac{a^2\cancel{\left(a+1\right)}\cancel{\left(a+x\right)}(a-1)\cancel{\left(a-x\right)}}{\cancel{\left(a+1\right)}\cancel{\left(a+x\right)}\cancel{\left(a-x\right)}}=}$${\displaystyle a^2(a-1)=}$${\displaystyle a^3-a^2}$;

тотожність доведено.