1098 Вправа 1098Алгебра 8 клас Бевз 2025 · § 21 · стор. 206

1098. Спрости вираз. а) (a + 4)/(a − 3) − (4a² − 1)/(2a² − 3a + 1) · (a² + 2a − 3)/(a² − 9); б) ((2a − 5)/(a + 2) − a/(a − 3) + 15/(a² − a − 6)) : (a − 10)/(a² + 2a); в) (x/(x² + 6x + 8) − 2/(x² + 3x + 2) + 6/(x² + 5x + 4)) · (x² + 2x)/(x − 3) − 3/(x − 3).

а) розкладаємо на множники:

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

${\displaystyle a^2-9=}$${\displaystyle (a-3)(a+3)}$;

${\displaystyle 2a^2-3a+1}$: ${\displaystyle D=}$${\displaystyle {(-3)}^2-4\cdot 2\cdot 1=}$${\displaystyle 1}$;

${\displaystyle a_{1,2}=}$${\displaystyle \frac{3\pm 1}{4}}$;

${\displaystyle a_1=}$${\displaystyle 1}$;

${\displaystyle a_2=}$${\displaystyle 0,5}$;

${\displaystyle 2a^2-3a+1=}$${\displaystyle 2(a-1)(a-0,5)=}$${\displaystyle \left(a-1\right)\left(2a-1\right)}$;

${\displaystyle a^2+2a-3}$: ${\displaystyle a_1+a_2=}$${\displaystyle -2}$;

${\displaystyle a_1{a}_2=}$${\displaystyle -3}$ → ${\displaystyle a_1=}$${\displaystyle 1,a_2=}$${\displaystyle -3}$;

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

1)

${\displaystyle \frac{4a^2-1}{2a^2-3a+1}\cdot \frac{a^2+2a-3}{a^2-9}=}$${\displaystyle \frac{(2a-1)(2a+1)}{(a-1)(2a-1)}\cdot \frac{(a+3)(a-1)}{(a-3)(a+3)}}$=

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

2)

${\displaystyle \frac{a+4}{a-3}-\frac{2a+1}{a-3}=}$${\displaystyle \frac{3-a}{a-3}=}$${\displaystyle -1}$;

б) ${\displaystyle a^2-a-6=}$${\displaystyle (a-3)(a+2)}$;

1)

${\displaystyle \frac{{2a-5}^{\mathrm{\setminus }a-3}}{a+2}-\frac{a^{\mathrm{\setminus }a+2}}{a-3}+\frac{15}{(a-3)(a+2)}}$=

=${\displaystyle \frac{2a^2-11a+15-a^2-2a+15}{(a-3)(a+2)}=}$${\displaystyle \frac{a^2-13a+30}{(a-3)(a+2)}}$=

=${\displaystyle \frac{\cancel{\left(a-3\right)}(a-10)}{\cancel{\left(a-3\right)}(a+2)}=}$${\displaystyle \frac{a-10}{a+2}}$;

2)

${\displaystyle \frac{a-10}{a+2}:\frac{a-10}{a^2+2a}=}$${\displaystyle \frac{a-10}{a+2}:\frac{a-10}{a(a+2)}=}$${\displaystyle \frac{a-10}{a+2}\cdot \frac{a(a+2)}{a-10}}$=

=${\displaystyle \frac{\cancel{\left(a-10\right)}\cdot a\cancel{\left(a+2\right)}}{\cancel{\left(a+2\right)}\cancel{\left(a-10\right)}}=}$${\displaystyle a}$;

в) знаменники: ${\displaystyle (x+2)(x+4)}$, ${\displaystyle (x+1)(x+2)}$, ${\displaystyle (x+1)(x+4)}$;

1)

${\displaystyle \frac{x^{\mathrm{\setminus }x+1}}{(x+2)(x+4)}-\frac{2^{\mathrm{\setminus }x+4}}{(x+1)(x+2)}+\frac{6^{\mathrm{\setminus }x+2}}{(x+1)(x+4)}}$=

=${\displaystyle \frac{x^2+x-2x-8+6x+12}{(x+1)(x+2)(x+4)}=}$${\displaystyle \frac{\cancel{\left(x+1\right)}\cancel{\left(x+4\right)}}{\cancel{\left(x+1\right)}(x+2)\cancel{\left(x+4\right)}}=}$${\displaystyle \frac{1}{x+2}}$;

2)

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

=${\displaystyle \frac{x}{x-3}-\frac{3}{x-3}=}$${\displaystyle \frac{x-3}{x-3}=}$${\displaystyle 1}$.