Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 169

а) a + b/a − b = $$ \frac{\frac{1}{1-x}+\frac{1}{1+x}}{\frac{1}{1-x}-\frac{1}{1+x}}=\frac{\frac{1-(1+x)}{(1-x) \cdot(1+x)}+\frac{1 \cdot(1-x)}{(1+x) \cdot(1-x)}}{\frac{1 \cdot(1+x)}{(1-x) \cdot(1+x)}-\frac{1+(1-x)}{(1+x) \cdot(1-x)}}=\frac{\frac{1+x}{1-x^2}+\frac{1-x}{1-x^2}}{\frac{1+x}{1-x^2}-\frac{1-x}{1-x^2}}=\frac{\frac{1+x+1-x}{1-x^2}}{\frac{1+x-1+x}{1-x^2}}= $$ $$ \frac{\frac{2}{1-x^2}}{\frac{2 x}{1-x^2}} $$ = 2/1 − x² : 2x/1 − x² = 2/1 − x² ⋅ 1 − x²/2x = 2 ⋅ (1 − x²)/(1 − x²) ⋅ 2x = 2 ⋅ (1 − x²) : 2(1 − x²)/(1 − x²) ⋅ 2x : 2(1 − x²) = 1/x
б) ax/a + x − bx/b − x = $$ \frac{a \cdot \frac{a b}{a-b}}{a+\frac{a b}{a-b}}-\frac{b \cdot \frac{a b}{a-b}}{b-\frac{a b}{a-b}}=\frac{\frac{a \cdot a b}{a-b}}{\frac{a}{1}+\frac{a b}{a-b}}-\frac{\frac{b \cdot a b}{a-b}}{\frac{b}{1}-\frac{a b}{a-b}}= $$ $$ \frac{\frac{a^2 b}{a-b}}{\frac{a(a-b)}{1-(a-b)}+\frac{a b}{a-b}}-\frac{\frac{a b^2}{a-b}}{\frac{b \cdot(a-b)}{1 \cdot(a-b)}-\frac{a b}{a-b}}=\frac{\frac{a^2 b}{a-b}}{\frac{a^2-a b+a b}{a-b}}-\frac{\frac{a b^2}{a-b}}{\frac{a b-b^2-a b}{a-b}}= $$ $$ \frac{\frac{a^2 b}{a-b}}{\frac{a^2}{a-b}-\frac{\frac{a b^2}{a-b}}{a-b}}=\frac{\frac{a^2 b}{a-b}}{\frac{a^2}{a-b}}+\frac{\frac{a b^2}{a-b}}{\frac{b^2}{a-b}} $$ = a²b/a − b : a²/a − b + ab²/a − b : b²/a − b = a²b/a − b ⋅ a − b/a² + ab²/a − b ⋅ a − b/b² = a²b ⋅ (a − b)/(a − b) ⋅ a² + ab² ⋅ (a − b)/(a − b) ⋅ b² = a²b ⋅ (a − b) : a²(a − b)/(a − b) ⋅ a² : a²(a − b) + ab² ⋅ (a − b) : b²(a − b)/(a − b) ⋅ b² : b²(a − b) = b/1 + a/1 = a + b