Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1287

a₁ + b₁$$\sqrt{2}$$, a₂ + b₂$$\sqrt{2}$$
Сумма равна a₁ + b₁$$\sqrt{2}$$ + a₂ + b₂$$\sqrt{2}$$ = (a₁ + a₂) + (b₁ + b₂)$$\sqrt{2}$$ a + b$$\sqrt{2}$$ a = a₁ + a₂, b = b₁ + b₂
Разность равна a₁ + b₁$$\sqrt{2}$$ − a₂ − b₂$$\sqrt{2}$$ = (a₁ − a₂) + (b₁ − b₂)$$\sqrt{2}$$ a + b$$\sqrt{2}$$ a = a₁ − a₂, b = b₁ − b₂
Произведение равно (a₁ + b₁$$\sqrt{2}$$)(a₂ + b₂$$\sqrt{2}$$) = a₁a₂ + a₁b₂$$\sqrt{2}$$ + a₂b₁$$\sqrt{2}$$ + 2b₁b₂ = a₁a₂ + (a₁b₂ + a₂b₁)$$\sqrt{2}$$ + 2b₁b₂ = (a₁a₂ + 2b₁b₂) + (a₁b₂ + a₂b₁)$$\sqrt{2}$$ a + b$$\sqrt{2}$$ a = a₁a₂ + 2b₁b₂, b = a₁b₂ + a₂b₁
Частное равно $$ \frac{a_1+b_1 \sqrt{2}}{a_2+b_2 \sqrt{2}}=\frac{\left(a_1+b_1 \sqrt{2}\right)\left(a_2-b_2 \sqrt{2}\right)}{\left(a_2+b_2 \sqrt{2}\right)\left(a_2-b_2 \sqrt{2}\right)} $$ = $$ \frac{a_1 a_2-a_1 b_2 \sqrt{2}+a_2 b_1 \sqrt{2}-2 b_1 b_2}{a_2^2-2 b_2^2}= $$ $$ \frac{a_1 a_2+\left(a_2 b_1-a_1 b_2\right) \sqrt{2}-2 b_1 b_2}{a_2^2-2 b_2^2} $$ = (a₁a₂ − 2b₁b₂/a₂² − 2b₂²) + (a₂b₁ − a₁b₂/a₂² − 2b₂²)$$\sqrt{2}$$ a + b$$\sqrt{2}$$ a = a₁a₂ − 2b₁b₂/a₂² − 2b₂²; b = a₁b₂ − a₂b₁/a₂² − 2b₂²
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