Question

помогите решить алгебру

Answer 1

Ответ:

$\boxed{\ a^2-b^2=(a-b)(a+b)\ ,\ (a\pm b)^2=a^2\pm 2ab+b^2\ }$

$4)\ \ (6-\sqrt5)(2+7\sqrt5)=12+42\sqrt5-2\sqrt5-35=40\sqrt5-23\\\\(2\sqrt{x} -5\sqrt{y})(2\sqrt{x}+5\sqrt{y})=(2\sqrt{x})^2-(5\sqrt{y})^2=4x-25y\\\\(5-\sqrt2)^2+(3+\sqrt2)^2=25-10\sqrt2+2+9+6\sqrt2+2=38-4\sqrt2\\\\(\sqrt{9-4\sqrt5}-\sqrt{9+4\sqrt5})^2=(9-4\sqrt5)-2\sqrt{(9-4\sqrt5)(9+4\sqrt5)}+(9+4\sqrt5)=\\\\=18-2\sqrt{81-80}=18-2\sqrt{1}=18-2=16$

$\displaystyle 5)\ \ \frac{x^2-13}{x-\sqrt{13}}=\frac{(x-\sqrt{13})(x+\sqrt{13})}{x-\sqrt{13}}=x+\sqrt{13}\\\\\\\frac{b-5\sqrt{b}}{b-25}=\frac{\sqrt{b}\cdot (\sqrt{b}-5)}{(\sqrt{b}-5)(\sqrt{b}+5)}=\frac{\sqrt{b}}{\sqrt{b}+5}\\\\\\\frac{x+16\sqrt{x}+64}{x-64}=\frac{(\sqrt{x}+8)^2}{(\sqrt{x}-8)(\sqrt{x}+8)}=\frac{\sqrt{x}+8}{\sqrt{x}-8}$

$\displaystyle 6)\ \ \frac{42}{5\sqrt7}=\frac{6\cdot 7}{5\sqrt7}=\frac{6\cdot (\sqrt7)^2}{5\sqrt7}=\frac{6\sqrt7}{5}\\\\\\\frac{14}{\sqrt{17}+\sqrt3}=\frac{14(\sqrt{17}-\sqrt3)}{(\sqrt{17}+\sqrt3)(\sqrt{17}-\sqrt3)}=\frac{14(\sqrt{17}-\sqrt3)}{17-3}=\sqrt{17}-\sqrt3$