Question

помогите пожалуйста!!!

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Answer 1

Ответ:

$\displaystyle 1)\ \ \dfrac{b^{\sqrt5}+c^{\sqrt7}}{b^{\sqrt{20}}+2\, b^{\sqrt5}c^{\sqrt7}+c^{\sqrt{28}}}=\dfrac{b^{\sqrt5}+c^{\sqrt{7}}}{b^{\sqrt{4\cdot 5}}+2\, b^{\sqrt5}c^{\sqrt7}+c^{\sqrt{4\cdot 7}}}=\\\\\\=\dfrac{b^{\sqrt5}+c^{\sqrt7}}{b^{2\sqrt{5}}+2\, b^{\sqrt5}c^{\sqrt7}+c^{2\sqrt{7}}}=\dfrac{b^{\sqrt5}+c^{\sqrt7}}{(b^{\sqrt{5}})^2+2\, b^{\sqrt5}c^{\sqrt7}+(c^{\sqrt{7}})^2}=$

$=\dfrac{b^{\sqrt5}+c^{\sqrt7}}{(b^{\sqrt{5}}+c^{\sqrt{7}})^2}=\dfrac{1}{b^{\sqrt5}+c^{\sqrt7}}$

$2)\ \ \displaystyle \frac{y^{\sqrt{18}}-27}{y^{\sqrt8}+3y^{\sqrt2}+9}=\frac{y^{\sqrt{9\cdot 2}}-3^3}{y^{\sqrt8}+3y^{\sqrt2}+9}=\frac{y^{3\sqrt2}-3^3}{y^{\sqrt8}+3y^{\sqrt2}+9}=\frac{(y^{\sqrt2})^3-3^3}{y^{\sqrt8}+3y^{\sqrt2}+9}=\\\\\\=\frac{(y^{\sqrt2}-3)((y^{\sqrt2})^2+3y^{\sqrt2}+3^2)}{y^{\sqrt8}+3y^{\sqrt2}+9}=\frac{(y^{\sqrt2}-3)(y^{2\sqrt2}+3y^{\sqrt2}+9)}{y^{\sqrt8}+3y^{\sqrt2}+9}=\\\\\\=\frac{(y^{\sqrt2}-3)(y^{\sqrt8}+3y^{\sqrt2}+9)}{y^{\sqrt8}+3y^{\sqrt2}+9}=y^{\sqrt2}-3$