Question

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Answer 1

$\displaystyle\bf3.\ \Bigg(-\dfrac{2}{b}\Bigg)^{-3}=\Bigg(-\frac{b}{2}\Bigg)^3=-\frac{b^3}{2^3}=-\frac{b^3}{8}\\\\\\\bf4.\ (4x-3)^2+18x=14x^2+5\\\\16x^2-24x+9+18x=14x^2+5\\\\16x^2-6x+9-14x^2-5=0\\\\2x^2-6x+4=0\ |\ \div2\\\\x^2-3x+2=0\\\\D=b^2-4ac=(-3)^2-4\cdot1\cdot2=9-8=1>0\Rightarrow2\ kornya\\\\x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{3\pm1}{2}\\\\x_1=\dfrac{3+1}{2}=2\\\\x_2=\dfrac{3-1}{2}=1$$5.\ \displaystyle\bf\Bigg(\dfrac{2a^2+10a+50}{a^3-125}-\dfrac{a+15}{a^2-25}\Bigg)\div\dfrac{a-10}{a^2+5a}=\\\\\\\displaystyle=\Bigg(\dfrac{2\big(a^2+5a+25\big)}{(a-5)(a^2+5a+25)}-\dfrac{a+15}{(a-5)(a+5)}\Bigg)\cdot\dfrac{a^2+5a}{a-10}=\\\\\\=\Bigg(\dfrac{2}{a-5}-\dfrac{a+15}{(a-5)(a+5)}\Bigg)\cdot\dfrac{a(a+5)}{a-10}=\\\\\\\dfrac{2(a+5)-(a+15)}{(a-5)(a+5)}\cdot\dfrac{a(a+5)}{a-10}=\dfrac{2a+10-a-15}{(a-5)(a+5)}\cdot\dfrac{a(a+5)}{a-10}=\\\\\\=\dfrac{a-5}{(a-5)(a+5)}\cdot\dfrac{a(a+5)}{a-10}=$

=   $\bf\dfrac{1}{a+5}\cdot\dfrac{a(a+5)}{a-10}=\dfrac{a}{a-10}$