Страницы 43-44 номер 2.3.21, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Найти числовое значение выражения: 1) \(\frac{100m^2-81n^2}{9n-10m}\) при \(m=-2{,}7\), \(n=3{,}2\); 2) \(\frac{0{,}7a-0{,}6b}{0{,}36b^2-0{,}49a^2}\) при \(a=40\), \(b=-30\); 3) \(\frac{0{,}008x^3-0{,}027y^3}{0{,}04x^2+0{,}06xy+0{,}09y^2}\) при \(x=\frac{5}{7}\), \(y=2\frac{2}{9}\); 4) \(\frac{\frac{y^2}{4}-\frac{x^2}{9}}{\frac{x^3}{27}+\frac{y^3}{8}}\) при \(x=-1{,}6\), \(y=-1{,}8\).
1) \[\frac{100m^2 - 81n^2}{9n - 10m} = \frac{(10m - 9n)(10m + 9n)}{-(10m - 9n)} = -(10m + 9n).\]
При \(m = -2{,}7\), \(n = 3{,}2\) получаем \(-(10 \cdot (-2{,}7) + 9 \cdot 3{,}2) = -(-27 + 28{,}8) = -1{,}8\).
2) \[\begin{aligned} &\frac{0{,}7a - 0{,}6b}{0{,}36b^2 - 0{,}49a^2} = \frac{-(0{,}6b - 0{,}7a)}{(0{,}6b - 0{,}7a)(0{,}6b + 0{,}7a)} = {} \\ &= -\frac{1}{0{,}6b + 0{,}7a}. \end{aligned}\]
При \(a = 40\), \(b = -30\) получаем \(-\frac{1}{0{,}6 \cdot (-30) + 0{,}7 \cdot 40} = -\frac{1}{-18 + 28} = -\frac{1}{10} = -0{,}1\).
3) \[\begin{aligned} &\frac{0{,}008x^3 - 0{,}027y^3}{0{,}04x^2 + 0{,}06xy + 0{,}09y^2} = {} \\ &= \frac{(0{,}2x - 0{,}3y)(0{,}04x^2 + 0{,}06xy + 0{,}09y^2)}{0{,}04x^2 + 0{,}06xy + 0{,}09y^2} = 0{,}2x - 0{,}3y. \end{aligned}\]
При \(x = \frac{5}{7}\), \(y = 2\frac{2}{9} = \frac{20}{9}\) получаем \(0{,}2 \cdot \frac{5}{7} - 0{,}3 \cdot \frac{20}{9} = \frac{1}{7} - \frac{2}{3} = \frac{3 - 14}{21} = -\frac{11}{21}\).
4) \[\begin{aligned} &\frac{\frac{y^2}{4} - \frac{x^2}{9}}{\frac{x^3}{27} + \frac{y^3}{8}} = \frac{\left(\frac{y}{2} - \frac{x}{3}\right)\left(\frac{y}{2} + \frac{x}{3}\right)}{\left(\frac{x}{3} + \frac{y}{2}\right)\left(\frac{x^2}{9} - \frac{xy}{6} + \frac{y^2}{4}\right)} = {} \\ &= \frac{\frac{y}{2} - \frac{x}{3}}{\frac{x^2}{9} - \frac{xy}{6} + \frac{y^2}{4}}. \end{aligned}\]
При \(x = -1{,}6\), \(y = -1{,}8\) имеем \(\frac{y}{2} = -0{,}9\) и \(\frac{x}{3} = -\frac{8}{15}\), поэтому числитель равен \(-0{,}9 + \frac{8}{15} = -\frac{27}{30} + \frac{16}{30} = -\frac{11}{30}\).
Далее \(\frac{x^2}{9} = \frac{2{,}56}{9} = \frac{64}{225}\), \(\frac{xy}{6} = \frac{2{,}88}{6} = 0{,}48\), \(\frac{y^2}{4} = \frac{3{,}24}{4} = 0{,}81\), поэтому знаменатель равен \(\frac{64}{225} - 0{,}48 + 0{,}81 = \frac{256}{900} - \frac{432}{900} + \frac{729}{900} = \frac{553}{900}\).
Значение выражения равно \(-\frac{11}{30} : \frac{553}{900} = -\frac{11}{30} \cdot \frac{900}{553} = -\frac{11 \cdot 30}{553} = -\frac{330}{553}\).
Ответ: 1) \(-1{,}8\); 2) \(-0{,}1\); 3) \(-\frac{11}{21}\); 4) \(-\frac{330}{553}\).
