Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 803

а) $$ \frac{x \sqrt{3}+\sqrt{2}}{x \sqrt{3}-\sqrt{2}}+\frac{x \sqrt{3}-\sqrt{2}}{x \sqrt{3}+\sqrt{2}}=\frac{10 x}{3 x^2-2} $$
     $$ \frac{(x \sqrt{3}+\sqrt{2})(x \sqrt{3}+\sqrt{2})}{(x \sqrt{3}-\sqrt{2})(x \sqrt{3}+\sqrt{2})}+\frac{(x \sqrt{3}-\sqrt{2})(x \sqrt{3}-\sqrt{2})}{(x \sqrt{3}+\sqrt{2})(x \sqrt{3}-\sqrt{2})}=\frac{10 x}{3 x^2-2} $$
     $$ \frac{(x \sqrt{3}+\sqrt{2})^2+(x \sqrt{3}-\sqrt{2})^2-10 x}{3 x^2-2}=0 $$
     $$ \frac{3 x^2+2 \sqrt{6}+2+3 x^2-2 \sqrt{6}+2-10 x}{3 x^2-2}=0 $$
     6x² − 10x + 4/3х² − 2 = 0      ОДЗ: x ≠ ±$$ \sqrt{\frac{2}{3}} $$      6x² − 10x + 4 = 0 : 2      3x² − 15x + 2 = 0      D = b² − 4ac = (−5)² − 4 · 3 · 2 = 25 − 24 = 1
     x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{5+\sqrt{1}}{2 · 3} $$ = 5 + 1/6 = 6/6 = 1
     x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{5-\sqrt{1}}{2 · 3} $$ = 5 − 1/6 = 4/6 = 2/3
     x₁ = 1, x₂ = 2/3
б) $$ \frac{1-y \sqrt{5}}{1+y \sqrt{5}}+\frac{1+y \sqrt{5}}{1-y \sqrt{5}}=\frac{9 y}{1-5 y^2} $$
     $$ \frac{(1-y \sqrt{5})(1-y \sqrt{5})}{(1+y \sqrt{5})(1-y \sqrt{5})}+\frac{(1+y \sqrt{5})(1+y \sqrt{5})}{(1-y \sqrt{5})(1+y \sqrt{5})}=\frac{9 y}{1-5 y^2} $$
     $$ \frac{(1-y \sqrt{5})^2+(1+y \sqrt{5})^2-9 y}{1-5 y^2}=0 $$
     $$ \frac{1-2 y \sqrt{5}+5 y^2+1+2 y \sqrt{5}+5 y^2-9 y}{1-5 y^2}=0 $$
     10y² − 9y + 2/1 − 5y² = 0      ОДЗ: y ≠ ±$$\sqrt{0,2}$$      10y² − 9y + 2 = 0      D = b² − 4ac = (−9)² − 4 · 10 · 2 = 81 − 80 = 1
     y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{9+\sqrt{1}}{2 · 10}= $$ = 9 + 1/20 = 10/20 = 0,5
     y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{9-\sqrt{1}}{2 · 10}= $$ = 9 − 1/20 = 8/20 = 0,4
     y₁ = 0,5, y₂ = 0,4