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$\displaystyle4).. x^{2}+ \frac{49}{ x^{2}}=50 \\ \\ x^{2}+( \frac{7}{x})^{2}=50 \\ \\x^{2}-2x* \frac{7}{x}+( \frac{7}{x})^{2}=50-2x* \frac{7}{x} \\ \\ (x- \frac{7}{x})^{2}=50-14 \\ \\ (x- \frac{7}{x})^{2}=36 \\ \\x- \frac{7}{x}=6 \\ \\ x- \frac{7}{x}=-6$

Ответ: {-6; 6}

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

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