Ответ:
Домножаем знаменатель и числитель на выражение, сопряжённое знаменателю, и применяем формулу разности квадратов
[tex]\bf (a-b)(a+b)=a^2-b^2[/tex] .
[tex]\displaystyle 4.16.\ \ \ \frac{5-a}{1+\sqrt5}=\frac{(5-a)(1-\sqrt5)}{(1+\sqrt5)(1-\sqrt5)}=\frac{5-5\sqrt5-a+a\sqrt5}{1-5}=-\frac{5-a-5\sqrt5+a\sqrt5}{4}\\\\\\\frac{7a+\sqrt{a}}{2-\sqrt{a}}=\frac{(7a+\sqrt{a})(2+\sqrt{a})}{(2-\sqrt{a})(2+\sqrt{a})}=\frac{14a+7a\sqrt{a}+2\sqrt{a}+a}{4-a}=\frac{15a+(7a+2)\sqrt{a}}{4-a}\\\\\\\frac{1,5c}{\sqrt3+c}=\frac{1,5c(\sqrt3-c)}{(\sqrt3+c)(\sqrt3-c)}=\frac{1,5c(\sqrt3-c)}{3-c^2}=\frac{1,5\sqrt3\, c-1,5c^2}{3-c^2}[/tex]
[tex]\displaystyle \frac{5c-1}{\sqrt{5c}-1}=\frac{(5c-1)(\sqrt{5c}+1)}{(\sqrt{5c}-1)(\sqrt{5c}+1)}=\frac{5c\sqrt{5c}+5c-\sqrt{5c}-1}{5c-1}=\frac{(5c-1)\sqrt{5c}+5c-1}{5c-1}=\\\\=\frac{(5c-1)(\sqrt{5c}+1)}{\sqrt{5c}-1}=\sqrt{5c}+1\\\\\\ili\ \ \ \ \frac{5c-1}{\sqrt{5c}-1}=\frac{(\sqrt{5c}-1)(\sqrt{5c}+1)}{\sqrt{5c}-1}=\sqrt{5c}+1\\\\\\\frac{3x+a}{\sqrt{5x}-\sqrt{a}}=\frac{(3x+a)(\sqrt{5x}+\sqrt{a})}{\sqrt{5x}-\sqrt{a})(\sqrt{5x}+\sqrt{a})}=\frac{3x\sqrt{5x}+3x\sqrt{a}+a\sqrt{5x}+a\sqrt{a}}{5x-a}[/tex]
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Ответ:
Домножаем знаменатель и числитель на выражение, сопряжённое знаменателю, и применяем формулу разности квадратов
[tex]\bf (a-b)(a+b)=a^2-b^2[/tex] .
[tex]\displaystyle 4.16.\ \ \ \frac{5-a}{1+\sqrt5}=\frac{(5-a)(1-\sqrt5)}{(1+\sqrt5)(1-\sqrt5)}=\frac{5-5\sqrt5-a+a\sqrt5}{1-5}=-\frac{5-a-5\sqrt5+a\sqrt5}{4}\\\\\\\frac{7a+\sqrt{a}}{2-\sqrt{a}}=\frac{(7a+\sqrt{a})(2+\sqrt{a})}{(2-\sqrt{a})(2+\sqrt{a})}=\frac{14a+7a\sqrt{a}+2\sqrt{a}+a}{4-a}=\frac{15a+(7a+2)\sqrt{a}}{4-a}\\\\\\\frac{1,5c}{\sqrt3+c}=\frac{1,5c(\sqrt3-c)}{(\sqrt3+c)(\sqrt3-c)}=\frac{1,5c(\sqrt3-c)}{3-c^2}=\frac{1,5\sqrt3\, c-1,5c^2}{3-c^2}[/tex]
[tex]\displaystyle \frac{5c-1}{\sqrt{5c}-1}=\frac{(5c-1)(\sqrt{5c}+1)}{(\sqrt{5c}-1)(\sqrt{5c}+1)}=\frac{5c\sqrt{5c}+5c-\sqrt{5c}-1}{5c-1}=\frac{(5c-1)\sqrt{5c}+5c-1}{5c-1}=\\\\=\frac{(5c-1)(\sqrt{5c}+1)}{\sqrt{5c}-1}=\sqrt{5c}+1\\\\\\ili\ \ \ \ \frac{5c-1}{\sqrt{5c}-1}=\frac{(\sqrt{5c}-1)(\sqrt{5c}+1)}{\sqrt{5c}-1}=\sqrt{5c}+1\\\\\\\frac{3x+a}{\sqrt{5x}-\sqrt{a}}=\frac{(3x+a)(\sqrt{5x}+\sqrt{a})}{\sqrt{5x}-\sqrt{a})(\sqrt{5x}+\sqrt{a})}=\frac{3x\sqrt{5x}+3x\sqrt{a}+a\sqrt{5x}+a\sqrt{a}}{5x-a}[/tex]