Помогите!Решить: 1)64х²-1/ х²-4 · (х+2)²/ х²-4 · (х-2)²/ 8х+1 2)х-6/ х²+6х+9 · х²+4х+4/ (х²+2)(х-2) · х³-9х/ (х-6)(х+2). Created by klimenkol21. algebra-ru

(х-6)(х+2)...

klimenkol21 @klimenkol21

August 2022 1 26 Report

Помогите!Решить:
1)64х²-1/ х²-4 · (х+2)²/ х²-4 · (х-2)²/ 8х+1
2)х-6/ х²+6х+9 · х²+4х+4/ (х²+2)(х-2) · х³-9х/ (х-6)(х+2)

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inblu

Verified answer

$\frac{64x^2-1}{x^2-4}* \frac{(x+2)^2}{x^2-4}* \frac{(x-2)^2}{8x+1}= \frac{(8x-1)(8x+1)(x+2)^2(x-2)^2}{(x-2)(x+2)(x-2)(x+2)(8x+1)}=8x-1 \\ \\ \frac{x-6}{x^2+6x+9}* \frac{x^2+4x+4}{(x^2+2)(x-2)}* \frac{x^3-9x}{(x-6)(x+2)}= \frac{(x-6)(x+2)^2*x*(x^2-9)}{(x+3)^2*(x^2+2)(x-2)(x-6)(x+2)}= \\ =\frac{(x+2)*x*(x-3)(x+3)}{(x+3)^2*(x^2+2)(x-2)}=\frac{x(x+2)(x-3)}{(x+3)*(x^2+2)(x-2)}$

1 votes Thanks 1

klimenkol21 спасибо большое

inblu да, не за что

osnovaniya trapecii 6sm i 9sm vysota 10sm najti rasstoyaniya ot tochki peresecheniya0c50ad104f6e89a3291d70d1402d398b 22936

Answer

klimenkol21 August 2022 | 0 Ответы

osnovaniya trapecii 6sm i 9sm vysota 10sm najti rasstoyaniya ot tochki peresecheniyadea6b551016c04c8383f618917441f22 11246

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Answers & Comments

Verified answer

2x-2...

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2x-2...

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osnovaniya trapecii 6sm i 9sm vysota 10sm najti rasstoyaniya ot tochki peresecheniya0c50ad104f6e89a3291d70d1402d398b 22936

osnovaniya trapecii 6sm i 9sm vysota 10sm najti rasstoyaniya ot tochki peresecheniyadea6b551016c04c8383f618917441f22 11246

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