Question

помогите пожалуйста решить

Answer 1

Ответ:

4) $\displaystyle \left(\frac{a^2}{a+5}-\frac{a^3}{a^2+10a+25}\right):\left(\frac{a}{a+5}-\frac{a^2}{a^2-25}\right)=\frac{5a-a^2}{a+5}$

5) $\displaystyle -13;\ 13$

Объяснение:

4)

$\displaystyle \left(\frac{a^2}{a+5}-\frac{a^3}{a^2+10a+25}\right):\left(\frac{a}{a+5}-\frac{a^2}{a^2-25}\right)=\\\\\\\left(\frac{a^2}{a+5}-\frac{a^3}{(a+5)^2}\right):\left(\frac{a}{a+5}-\frac{a^2}{(a+5)(a-5)}\right)=\\\\\\\left(\frac{a^2(a+5)}{(a+5)^2}-\frac{a^3}{(a+5)^2}\right):\left(\frac{a(a-5)}{(a+5)(a-5)}-\frac{a^2}{(a+5)(a-5)}\right)=$

$\displaystyle \frac{a^2(a+5)-a^3}{(a+5)^2}:\frac{a(a-5)-a^2}{(a+5)(a-5)}=\frac{a^3+5a^2-a^3}{(a+5)^2}:\frac{a^2-5a-a^2}{(a+5)(a-5)}=\\\\\\\frac{5a^2}{(a+5)^2}:\frac{-5a}{(a+5)(a-5)}=\frac{5a^2}{(a+5)^2}\cdot \frac{(a+5)(a-5)}{-5a}=\\\\\\\frac{a}{a+5}\cdot \frac{a-5}{-1}=\frac{-a(a-5)}{a+5}=\frac{-a^2+5a}{a+5}=\frac{5a-a^2}{a+5}$

5)

$\displaystyle 16x^2+\frac{9}{x^2}=145\\\\\\(4x)^2+\left(\frac{3}{x}\right)^2=145\\\\\\(4x)^2+2\cdot 4x \cdot \frac{3}{x}+\left(\frac{3}{x}\right)^2-2\cdot 4x \cdot \frac{3}{x}=145\\\\\\\left(4x+\frac{3}{x}\right)^2-24=145\\\\\\\left(4x+\frac{3}{x}\right)^2=145+14\\\\\\\left(4x+\frac{3}{x}\right)^2=169\\\\\\4x+\frac{3}{x}=-\sqrt{169}\ \ \ \ \ \ \ \ \ \ 4x+\frac{3}{x}=\sqrt{169}\\\\\\4x+\frac{3}{x}=-13\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4x+\frac{3}{x}=13$