Алгебра: Берілген өрнектердің тепе-тең болатынын дәлелдеңдер

Объяснение:1)2)3)4)...

Берілген өрнектердің тепе-тең болатынын дәлелдеңдер $1) \frac{1}{6x + 10} - \frac{1}{9x - 15} + \frac{5}{9 {x}^{2} - 25 } \: \frac{1}{6.(3x - 5)}$

$2) \frac{1}{2x - 8} + \frac{1}{40 - 10x} + \frac{1}{ {x}^{2} - 8x + 16 } \: \: \frac{2x - 3}{5(x - 4 {)}^{2} }$
$3) \frac{1}{x - 2} + \frac{x - 2}{x {}^{2} + 2x + 4 } - \frac{6x}{ {x}^{3} - 8} \: \: \frac{2x - 4}{ {x}^{2} + 2x + 4}$
$4) \frac{2a {}^{2} + 7a + 3 }{ {a}^{3} - 1} - \frac{1 - 2a}{ {a}^{2} + a + 1 } - \frac{1}{a - 1} \: \: \: \frac{3}{a - 1}$

Ответ:

Лена17617

15.10.2020 16:35

Объяснение:

$\frac{1}{6x+10}-\frac{1}{9x-15} +\frac{5}{9x^2-25}=\frac{1}{2*(3x+5)} -\frac{1}{3*(3x-5)}+\frac{5}{(3x-5)(3x+5) }=\\=\frac{3*(3x-5)-2*(3x+5)+5*2*3}{2*3*(3x-5)(3x+5)}=\frac{9x-15-6x-10+30}{6*(3x-5)(3x+5)} =\frac{3x+5}{6*(3x-5)(3x+5)}=\frac{1}{6*(3x-5)}.$

$\frac{1}{2x-8}+\frac{1}{40-10x} +\frac{1}{x^2-8x+16}=\frac{1}{2*(x-4)}+\frac{1}{10*(4-x)}+\frac{1}{(x-4)^2}=\\ =\frac{1}{2*(x-4)}-\frac{1}{10*(x-4)} +\frac{1}{(x-4)^2} =\frac{5*(x-4)-(x-4)+10}{10*(x-4)^2} =\frac{5x-20-x+4+10}{y} =\\=\frac{4x-6}{10*(x-4)^2} =\frac{2*(2x-3)}{10*(x-4)^2} =\frac{2x-3}{5*(x-4)^2} .$

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

$\frac{2a^2+7a+3}{a^3-1} -\frac{1-2a}{a^2+a+1} -\frac{1}{a-1}=\frac{2a^2+7a+3-(1-2a)(a-1)-(a^2+a+1)}{(a-1)(a^2+a+1)} =\\ =\frac{2a^2+7a+3-(a-2a^2-1+2a)-a^2-a-1}{(a-1)(a^2+a+1)} =\frac{2a^2+7a+3-3a+2a^2+1-a^2-a-1}{(a-1)(a^2+a+1)} =\\=\frac{3a^2+3a+3}{(a-1)(a^2+a+1)} =\frac{3*(a^2+a+1)}{(a-1)(a^2+a+1)} =\frac{3}{a-1}.$