Question

помогите пожалуйста очень срочно прошу вас умоляю решите все буду очень благодарна​

Answer 1

Ответ:

$\dfrac{2}{(y-x)(y-z)};$

$\dfrac{48}{1-a^{32}}=\dfrac{48}{1-a^{32}};$

Объяснение:

$\dfrac{1}{x^{2}-xz-xy+yz}+\dfrac{2}{y^{2}-xy-yz+xz}+\dfrac{1}{z^{2}-xz-yz+xy}=$

$=\dfrac{1}{x(x-z)-y(x-z)}+\dfrac{2}{y(y-x)-z(y-x)}+\dfrac{1}{x(z-x)-y(z-x)}=$

$=\dfrac{1}{(x-z)(x-y)}+\dfrac{2}{(y-x)(y-z)}+\dfrac{1}{(z-x)(x-y)}=\dfrac{1}{(x-z)(x-y)}-$

$-\dfrac{2}{(y-z)(x-y)}-\dfrac{1}{(x-z)(x-y)}=\dfrac{y-z-2(x-z)-(y-z)}{(x-z)(x-y)(y-z)}=$

$=\dfrac{y-z-2x+2z-y+z}{(x-z)(x-y)(y-z)}=\dfrac{-2x+2z}{(x-z)(x-y)(y-z)}=\dfrac{-2(x-z)}{(x-z)(x-y)(y-z)}=$

$=\dfrac{-2}{(x-y)(y-z)}=\dfrac{2}{-(x-y)(y-z)}=\dfrac{2}{(y-x)(y-z)};$

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$\dfrac{3}{1-a^{2}}+\dfrac{3}{1+a^{2}}+\dfrac{6}{1+a^{4}}+\dfrac{12}{1+a^{8}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{3(1+a^{2})+3(1-a^{2})}{(1-a^{2})(1+a^{2})}+\dfrac{6}{1+a^{4}}+\dfrac{12}{1+a^{8}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{3+3a^{2}+3-3a^{2}}{1^{2}-(a^{2})^{2}}+\dfrac{6}{1+a^{4}}+\dfrac{12}{1+a^{8}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{6}{1-a^{4}}+\dfrac{6}{1+a^{4}}+\dfrac{12}{1+a^{8}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{6(1+a^{4})+6(1-a^{4})}{(1-a^{4})(1+a^{4})}+\dfrac{12}{1+a^{8}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{6+6a^{4}+6-6a^{4}}{1^{2}-(a^{4})^{2}}+\dfrac{12}{1+a^{8}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{12}{1-a^{8}}+\dfrac{12}{1+a^{8}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{12(1+a^{8})+12(1-a^{8})}{(1-a^{8})(1+a^{8})}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{12+12a^{8}+12-12a^{8}}{1^{2}-(a^{8})^{2}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{24}{1-a^{16}}+\dfrac{24}{1+a^{16}}=\dfrac{48}{1-a^{32}};$

$\dfrac{24(1+a^{16})+24(1-a^{16})}{(1-a^{16})(1+a^{16})}=\dfrac{48}{1-a^{32}};$

$\dfrac{24+24a^{16}+24-24a^{16}}{1^{2}-(a^{16})^{2}}=\dfrac{48}{1-a^{32}};$

$\dfrac{48}{1-a^{32}}=\dfrac{48}{1-a^{32}};$

Тождество доказано.