Question

460,421 пожалуйста на листочке

Answer 1

№461
$\frac{a}{a^2-ab} - \frac{b}{a^2-b^2}= \frac{a}{a(a-b)} - \frac{b}{(a-b)(a+b)} = \frac{a^2+ab-ab}{a(a-b)(a+b)} = \frac{a^2}{a(a-b)(a+b)}$
Если $a=-1;b= \sqrt{3}$, то 
$\frac{(-1)^2}{-1(-1+ \sqrt{3} )(-1- \sqrt{3} )} = \frac{1}{-1(1-3)}= \frac{1}{-1*(-2)} = \frac{1}{2} =0.5$

№421
$\frac{n^3- \sqrt{2}*n^2 }{n^2-2} = \frac{n^2(n- \sqrt{2} )}{n^2-2}$
Если $n=2 \sqrt{2}$, то 
$\frac{(2 \sqrt{2})^2( 2\sqrt{2}- \sqrt{2} )}{(2 \sqrt{2} )^2-2} = \frac{8 \sqrt{2} }{8-2} = \frac{ \sqrt{64*2} }{6} = \frac{ \sqrt{128} }{ \sqrt{36} } = \sqrt{ \frac{128}{36} } = \sqrt{ \frac{2^6*3}{2^2*3^2} } = \sqrt \frac{2^4}{3} }$
Или вообще вместо $\frac{8 \sqrt{2} }{6} = \frac{8}{6} \sqrt{2} =1 \frac{1}{3} \sqrt{2}$

Answer 2

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