667 Вправа 667Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 195

667. Розв’яжіть рівняння:

1) ${\displaystyle (2x+1)(x+1)-3x=}$${\displaystyle 4x^2}$;

${\displaystyle 2x^2+2x+x+1-3x=}$${\displaystyle 4x^2}$;

${\displaystyle 2x^2+3x+1-3x-4x^2=}$${\displaystyle 0}$;

${\displaystyle -2x^2+1=}$${\displaystyle 0}$;

${\displaystyle -2x^2=}$${\displaystyle -1}$ | ${\displaystyle :\left(-2\right)}$

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

${\displaystyle x=}$${\displaystyle \sqrt{\frac{1}{2}}}$ або ${\displaystyle x=}$${\displaystyle -\sqrt{\frac{1}{2}}}$;

${\displaystyle x=}$${\displaystyle \frac{\sqrt{2}}{2}}$ або ${\displaystyle x=}$${\displaystyle -\frac{\sqrt{2}}{2}}$;

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

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

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

${\displaystyle -x^2=}$${\displaystyle 1}$ | ${\displaystyle :\left(-1\right)}$

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

коренів немає;

3) ${\displaystyle (5x+1)(x+2)-6x^2-11x=}$${\displaystyle 0}$;

${\displaystyle 5x^2+10x+x+2-6x^2-11x=}$${\displaystyle 0}$;

${\displaystyle -x^2+2=}$${\displaystyle 0}$;

${\displaystyle -x^2=}$${\displaystyle -2}$ | ${\displaystyle :\left(-1\right)}$

${\displaystyle x^2=}$${\displaystyle 2}$;

${\displaystyle x=}$${\displaystyle \sqrt{2}}$ або ${\displaystyle x=}$${\displaystyle -\sqrt{2}}$;

4) ${\displaystyle (-x+5)(x+4)=}$${\displaystyle x(x+1)}$;

${\displaystyle -x^2-4x+5x+20=}$${\displaystyle x^2+x}$;

${\displaystyle -x^2+x+20-x^2-x=}$${\displaystyle 0}$;

${\displaystyle -2x^2+20=}$${\displaystyle 0}$;

${\displaystyle -2x^2=}$${\displaystyle -20}$ | ${\displaystyle :\left(-2\right)}$

${\displaystyle x^2=}$${\displaystyle 10}$;

${\displaystyle x=}$${\displaystyle \sqrt{10}}$ або ${\displaystyle x=}$${\displaystyle -\sqrt{10}}$.