211 Вправа 211Алгебра 8 клас Бевз 2025 · § 4 · стор. 42

211. Доведи тотожність.

а) Додаємо послідовно, починаючи з перших двох дробів:

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

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

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

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

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

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

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

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

=${\displaystyle \frac{8}{(1-x)(x+1)(x^2+1)(x^4+1)}=}$${\displaystyle \frac{8}{1-x^8}}$;

отже, ${\displaystyle \frac{8}{1-x^8}=}$${\displaystyle \frac{8}{1-x^8}}$ — тотожність доведено;

б) так само:

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

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

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

=${\displaystyle \frac{2(4x^2+1)+2(1-2x)(2x+1)}{(1-2x)(2x+1)(4x^2+1)}=}$${\displaystyle \frac{8x^2+2+2-8x^2}{(1-2x)(2x+1)(4x^2+1)}}$=

=${\displaystyle \frac{4}{(1-2x)(2x+1)(4x^2+1)}=}$${\displaystyle \frac{4}{1-16x^4}}$;

${\displaystyle \frac{4}{1-16x^4}+\frac{4}{1+16x^4}}$=

=${\displaystyle \frac{4^{\mathrm{\setminus }16x^4+1}}{(1-2x)(2x+1)(4x^2+1)}+\frac{4^{\mathrm{\setminus }\left(1-2x\right)\left(2x+1\right)\left(4x^2+1\right)}}{16x^4+1}}$=

=${\displaystyle \frac{4(16x^4+1)+4(1-2x)(2x+1)(4x^2+1)}{(1-2x)(2x+1)(4x^2+1)(16x^4+1)}}$=

=${\displaystyle \frac{64x^4+4+4-64x^4}{(1-2x)(2x+1)(4x^2+1)(16x^4+1)}}$=

=${\displaystyle \frac{8}{(1-2x)(2x+1)(4x^2+1)(16x^4+1)}=}$${\displaystyle \frac{8}{1-256x^8}}$;

отже, ${\displaystyle \frac{8}{1-256x^8}=}$${\displaystyle \frac{8}{1-256x^8}}$ — тотожність доведено.