996 Вправа 996Алгебра 8 клас Бевз 2025 · § 19 · стор. 186

996. Розв’яжи рівняння, використовуючи метод заміни змінної. а) (x + 3)⁴ + (x + 3)² − 2 = 0; б) (x² − 3x)² − 2(x² − 3x) − 8 = 0; в) ((2x + 1)/x)² − 6·(2x + 1)/x = −5; г) ((√x − 1)/2)² − (√x − 1)/2 = 12; ґ) x² + 2x + 2(x + 1) − 23 = 0; д) x² + 5 = 2√(x² + 5) + 3.

а) ${\displaystyle {(x+3)}^4+{(x+3)}^2-2=}$${\displaystyle 0}$;

заміна ${\displaystyle {(x+3)}^2=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 1,c=}$${\displaystyle -2}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 1^2-4\cdot 1\cdot (-2)=}$${\displaystyle 1+8=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{9}}{2}=}$${\displaystyle \frac{-1\pm 3}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-1+3}{2}=}$${\displaystyle 1}$;

${\displaystyle t_2=}$${\displaystyle \frac{-1-3}{2}=}$${\displaystyle -2}$;

${\displaystyle t=}$${\displaystyle -2<0}$ — не підходить;

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

${\displaystyle x+3=}$${\displaystyle \pm 1}$;

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

${\displaystyle x=}$${\displaystyle -4}$;

б) ${\displaystyle {(x^2-3x)}^2-2(x^2-3x)-8=}$${\displaystyle 0}$;

заміна ${\displaystyle x^2-3x=}$${\displaystyle t}$:

${\displaystyle t^2-2t-8=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -2,c=}$${\displaystyle -8}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-8)=}$${\displaystyle 4+32=}$${\displaystyle 36}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{2\pm \sqrt{36}}{2}=}$${\displaystyle \frac{2\pm 6}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{2+6}{2}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{2-6}{2}=}$${\displaystyle -2}$;

${\displaystyle x^2-3x=}$${\displaystyle 4}$:

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -3,c=}$${\displaystyle -4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-4)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{25}}{2}=}$${\displaystyle \frac{3\pm 5}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+5}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{3-5}{2}=}$${\displaystyle -1}$;

${\displaystyle x^2-3x=}$${\displaystyle -2}$:

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -3,c=}$${\displaystyle 2}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot 2=}$${\displaystyle 9-8=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{1}}{2}=}$${\displaystyle \frac{3\pm 1}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+1}{2}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{3-1}{2}=}$${\displaystyle 1}$;

в) ${\displaystyle {\left(\frac{2x+1}{x}\right)}^2-6\cdot \frac{2x+1}{x}=}$${\displaystyle -5}$;

заміна ${\displaystyle \frac{2x+1}{x}=}$${\displaystyle t}$, ${\displaystyle x\ne 0}$:

${\displaystyle t^2-6t=}$${\displaystyle -5}$;

${\displaystyle t^2-6t+5=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -6,c=}$${\displaystyle 5}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-6)}^2-4\cdot 1\cdot 5=}$${\displaystyle 36-20=}$${\displaystyle 16}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{6\pm \sqrt{16}}{2}=}$${\displaystyle \frac{6\pm 4}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{6+4}{2}=}$${\displaystyle 5}$;

${\displaystyle t_2=}$${\displaystyle \frac{6-4}{2}=}$${\displaystyle 1}$;

${\displaystyle \frac{2x+1}{x}=}$${\displaystyle 5}$;

${\displaystyle 2x+1=}$${\displaystyle 5x}$;

${\displaystyle 3x=}$${\displaystyle 1}$;

${\displaystyle x=}$${\displaystyle \frac{1}{3}}$;

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

${\displaystyle 2x+1=}$${\displaystyle x}$;

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

г) ${\displaystyle {\left(\frac{\sqrt{x}-1}{2}\right)}^2-\frac{\sqrt{x}-1}{2}=}$${\displaystyle 12}$;

заміна ${\displaystyle \frac{\sqrt{x}-1}{2}=}$${\displaystyle t}$:

${\displaystyle t^2-t=}$${\displaystyle 12}$;

${\displaystyle t^2-t-12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle -12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-12)=}$${\displaystyle 1+48=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{49}}{2}=}$${\displaystyle \frac{1\pm 7}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{1+7}{2}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{1-7}{2}=}$${\displaystyle -3}$;

${\displaystyle \frac{\sqrt{x}-1}{2}=}$${\displaystyle 4}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 9}$;

${\displaystyle x=}$${\displaystyle 81}$;

${\displaystyle \frac{\sqrt{x}-1}{2}=}$${\displaystyle -3}$;

${\displaystyle \sqrt{x}=}$${\displaystyle -5}$ — коренів немає;

ґ) ${\displaystyle x^2+2x+2(x+1)-23=}$${\displaystyle 0}$;

оскільки ${\displaystyle x^2+2x=}$${\displaystyle {(x+1)}^2-1}$, то

${\displaystyle {(x+1)}^2-1+2(x+1)-23=}$${\displaystyle 0}$;

заміна ${\displaystyle x+1=}$${\displaystyle t}$:

${\displaystyle t^2+2t-24=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 2,c=}$${\displaystyle -24}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 2^2-4\cdot 1\cdot (-24)=}$${\displaystyle 4+96=}$${\displaystyle 100}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-2\pm \sqrt{100}}{2}=}$${\displaystyle \frac{-2\pm 10}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-2+10}{2}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{-2-10}{2}=}$${\displaystyle -6}$;

${\displaystyle x+1=}$${\displaystyle 4}$;

${\displaystyle x=}$${\displaystyle 3}$;

${\displaystyle x+1=}$${\displaystyle -6}$;

${\displaystyle x=}$${\displaystyle -7}$;

д) ${\displaystyle x^2+5=}$${\displaystyle 2\sqrt{x^2+5}+3}$;

заміна ${\displaystyle \sqrt{x^2+5}=}$${\displaystyle t}$, ${\displaystyle t>0}$:

${\displaystyle t^2=}$${\displaystyle 2t+3}$;

${\displaystyle t^2-2t-3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -2,c=}$${\displaystyle -3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-3)=}$${\displaystyle 4+12=}$${\displaystyle 16}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{2\pm \sqrt{16}}{2}=}$${\displaystyle \frac{2\pm 4}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{2+4}{2}=}$${\displaystyle 3}$;

${\displaystyle t_2=}$${\displaystyle \frac{2-4}{2}=}$${\displaystyle -1}$;

${\displaystyle t=}$${\displaystyle -1<0}$ — не підходить;

${\displaystyle \sqrt{x^2+5}=}$${\displaystyle 3}$;

${\displaystyle x^2+5=}$${\displaystyle 9}$;

${\displaystyle x^2=}$${\displaystyle 4}$;

${\displaystyle x=}$${\displaystyle \pm 2}$.