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DakotaDp
11.10.2021 17:31
Алгебра
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2x в квадрате -8x -( x-1)(x-4)=2x в квадрате +2x-14x-14 ​

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Heda0101
Heda0101
4,7(55 оценок)

\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)};

\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.

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