8. 
Область определения:
1)
x \geq \frac{3}{4} " alt=" 4x - 3 \geq 0 => x \geq \frac{3}{4} " align="absmiddle" class="latex-formula">
2)
x \geq -1 " alt=" x + 1 \geq 0 => x \geq -1 " align="absmiddle" class="latex-formula">
3)
x \geq \frac{2}{5} " alt=" 5x - 2 \geq 0 => x \geq \frac{2}{5} " align="absmiddle" class="latex-formula">
Значит, 
Возведём в квадрат обе части уравнения и решим его:










Зная область определения, правильный ответ: 
9. 
Упростим выражение и получим:






Ответ: sin²α
10. 