Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 167

а) $$ \frac{2-\frac{a}{x}}{2+\frac{a}{x}}=\frac{\frac{2}{1}-\frac{a}{x}}{\frac{2}{1}+\frac{a}{x}}=\frac{\frac{2 \cdot x}{1 \cdot x}-\frac{a}{x}}{\frac{2 \cdot x}{1 \cdot x}+\frac{a}{x}}=\frac{\frac{2 x-a}{x}}{\frac{2 x+a}{x}} $$ = 2x − a/x : 2x + a/x = 2x − a/x ⋅ x/2x + a = (2x − a) ⋅ x/x ⋅ (2x + a) = (2x − a) ⋅ x : x/x ⋅ (2x + a) : x = 2x − a/2x + a
б) $$ \frac{\frac{a-b}{c}+3}{\frac{a+b}{c}-1}=\frac{\frac{a-b}{c}+\frac{3}{1}}{\frac{a+b}{c}-\frac{1}{1}}=\frac{\frac{a-b}{c}+\frac{3-c}{1-c}}{\frac{a+b}{c}-\frac{1-c}{1-c}}=\frac{\frac{a-b+3 c}{c}}{\frac{a+b-c}{c}} $$ = a − b + 3c/c : a + b − c/c = a − b + 3c/c ⋅ c/a + b − c = (a − b + 3c) ⋅ c/c ⋅ (a + b − c) = (a − b + 3c) ⋅ c : c/c ⋅ (a + b − c) : c = a − b + 3c/a + b − c
в)$$ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{1 \cdot y}{x \cdot y}+\frac{1 \cdot x}{y \cdot x}}{\frac{1 \cdot y}{x \cdot y}-\frac{1 \cdot x}{y \cdot x}}=\frac{\frac{y+x}{x y}}{\frac{y-x}{x y}} $$ = y + x/xy : y − x/xy = y + x/xy ⋅ xy/y − x = (y + x) ⋅ xy/xy ⋅ (y − x) = (y + x) ⋅ xy : xy/xy ⋅ (y − x) : xy = y + x/y − x
г) $$ \frac{x-y}{\frac{x}{y}-\frac{y}{x}}=\frac{x-y}{\frac{x \cdot x}{y \cdot x}-\frac{y \cdot y}{x \cdot y}}=\frac{x-y}{\frac{x^2-y^2}{x y}} $$ = (x − y) : x² − y²/xy = (x − y)/1 ⋅ xy/x² − y² = (x − y) ⋅ xy/1 ⋅ (x − y)(x + y) = (x − y) ⋅ xy : (x − y)/1 ⋅ (x − y)(x + y) : (x − y) = xy/x + y