Question

ПОМОГИТЕ СРОЧНО. НУЖНО РЕШИТЬ ЛЮБОЙ ИЗ ЭТИХ​

Answer 1

$1)\ \dfrac{2a}{4a^2-b^2}+\dfrac{1}{b-2a}=\dfrac{2a}{(2a-b)(2a+b)}+\dfrac{1}{-(2a-b)}=\\\\\\=\dfrac{2a}{(2a-b)(2a+b)}-\dfrac{1}{2a-b}=\dfrac{2a-(2a+b)}{(2a-b)(2a+b)}=\dfrac{2a-2a-b}{4a^2-b^2}=-\dfrac{b}{4a^2-b^2}$

$2)\ \bigg(\dfrac{1}{cos^2\alpha}-1\bigg)-\dfrac{sin^2\alpha-1}{1-cos^2\alpha}=\dfrac{1}{cos^2\alpha}-1-\dfrac{-(1-sin^2\alpha)}{sin^2\alpha}=\\\\\\=\dfrac{1}{cos^2\alpha}-1-\dfrac{-cos^2\alpha}{sin^2\alpha}=\dfrac{1}{cos^2\alpha}-1+\dfrac{cos^2\alpha}{sin^2\alpha}=\dfrac{sin^2\alpha-cos^2\alpha sin^2\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}=\\\\\\=\dfrac{sin^2\alpha(1-cos^2\alpha)+cos^4\alpha}{cos^2\alpha sin^2\alpha}=\dfrac{sin^2\alpha sin^2\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}=$

$=\dfrac{sin^4\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}$