Выполните сложение или вычитание дробей, Полученные выражения упрастите: а)
[tex] \frac{a}{4} + \frac{2a - b}{4} - \frac{b - 2a}{4} [/tex]
B)
[tex] \frac{ {a}^{2} }{a + 5 } - \frac{25}{a + 5} [/tex]
c)
[tex] \frac{1}{10c - 1} - \frac{ {100c}^{2} }{10c - 1} [/tex]
D)
[tex] \frac{7x}{x - y} - \frac{7y}{y - x} [/tex]
E)
[tex] \frac{ {d}^{2} }{d - 3} + \frac{9}{3 - d} [/tex]
F)
[tex] \frac{ {k}^{2} }{ {(k + 4)}^{2} } - \frac{16}{ { (- 4 - k)}^{2} } [/tex]
G)
[tex] \frac{ {9m}^{2} }{ {(m - n)}^{2} } - \frac{ {9n}^{2} }{ {(n - m)}^{2} } [/tex]
(Старался 10 минут писал вопрос)
[tex] \frac{a}{4} + \frac{2a - b}{4} - \frac{b - 2a}{4} [/tex]
B)
[tex] \frac{ {a}^{2} }{a + 5 } - \frac{25}{a + 5} [/tex]
c)
[tex] \frac{1}{10c - 1} - \frac{ {100c}^{2} }{10c - 1} [/tex]
D)
[tex] \frac{7x}{x - y} - \frac{7y}{y - x} [/tex]
E)
[tex] \frac{ {d}^{2} }{d - 3} + \frac{9}{3 - d} [/tex]
F)
[tex] \frac{ {k}^{2} }{ {(k + 4)}^{2} } - \frac{16}{ { (- 4 - k)}^{2} } [/tex]
G)
[tex] \frac{ {9m}^{2} }{ {(m - n)}^{2} } - \frac{ {9n}^{2} }{ {(n - m)}^{2} } [/tex]
(Старался 10 минут писал вопрос)
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Answers & Comments
Ответ:
Используем формулу разности квадратов [tex]a^2-b^2=(a-b)(a+b)[/tex] и то,
что [tex](m-n)^2=(n-m)^2[/tex] . Раскладываем числители и знаменатели дробей на множители, и если это возможно, сокращаем одинаковые множители .
[tex]\displaystyle a)\ \ \frac{a}{4}+\frac{2a-b}{4}-\frac{b-2a}{4}=\frac{a+2a-b-b+2a}{4}=\frac{5a-2b}{4}\\\\\\b)\ \ \frac{a^2}{a+5}-\frac{25}{a+5}=\frac{a^2-25}{a+5}=\frac{(a-5)(a+5)}{a+5}=a-5\\\\\\c)\ \ \frac{1}{10c-1}-\frac{100c^2}{10c-1}=\frac{1-100c^2}{10c-1}=\frac{(1-10c)(1+10c)}{-(1-10c)}=-(1+10c)=-1-10c\\\\\\d)\ \ \frac{7x}{x-y}-\frac{7y}{y-x}=\frac{7x}{x-y}+\frac{7y}{x-y}=\frac{7(x+y)}{x-y}[/tex]
[tex]\displaystyle e)\ \ \frac{d^2}{d-3}+\frac{9}{3-d}=\frac{d^2}{d-3}-\frac{9}{d-3}=\frac{d^2-9}{d-3}=\frac{(d-3)(d+3)}{d-3}=d+3\\\\\\f)\ \ \frac{k^2}{(k+4)^2}-\frac{16}{(-4-k)^2}=\frac{k^2}{(k+4)^2}-\frac{16}{(k+4)^2}=\frac{k^2-16}{(k+4)^2}=\frac{(k-4)(k+4)}{(k+4)^2}=\\\\\\=\frac{k-4}{k+4}\\\\\\g)\ \ \frac{9m^2}{(m-n)^2}-\frac{9n^2}{(n-m)^2}=\frac{9m^2}{(m-n)^2}-\frac{9n^2}{(m-n)^2}=\frac{9(m^2-n^2)}{(m-n)^2}=\\\\\\=\frac{9(m-n)(m+n)}{(m-n)^2}=\frac{9(m+n)}{m-n}[/tex]
Verified answer
[tex] \frac{a}{4} + \frac{2a - b}{4} - \frac{b - 2a}{4} = \frac{a + 2a - b - b + 2a}{4} = \frac{5a - 2b}{4} [/tex]
[tex] \frac{ {a}^{2} }{a + 5} - \frac{25}{a + 5} = \frac{ {a}^{2} - 25 }{a + 5} = \frac{(a - 5)(a + 5)}{a + 5} = a - 5[/tex]
[tex] \frac{1}{10c - 1} - \frac{100 {c}^{2} }{10c - 1} = \frac{1 - 100 {c}^{2} }{10c - 1} = \frac{(1 - 10c)(1 + 10c) }{10c - 1} = - \frac{(10c - 1)(10c + 1)}{10c - 1} = - 10c - 1[/tex]
[tex] \frac{7x}{x - y} - \frac{7y}{y - x} = \frac{7x}{x - y} + \frac{7y}{x - y} = \frac{7x + 7y}{x - y} [/tex]
[tex] \frac{ {d}^{2} }{d - 3} + \frac{9}{3 - d} = \frac{ {d}^{2} }{d - 3} - \frac{9}{d - 3} = \frac{ {d}^{2} - 9}{d - 3} = \frac{(d - 3)(d + 3)}{d - 3} = d + 3[/tex]
[tex] \frac{ {k}^{2} }{ {(k + 4)}^{2} } - \frac{16}{ {( - 4 - k)}^{2} } = \frac{ {k}^{2} }{ {(k + 4)}^{2} } - \frac{16}{ {(k + 4)}^{2} } = \frac{ {k}^{2} - 16 }{ {(k + 4)}^{2} } = \frac{(k - 4)(k + 4)}{ {(k + 4)}^{2} } = \frac{k - 4}{k + 4} [/tex]
[tex] \frac{9 {m}^{2} }{ {(m - n)}^{2} } - \frac{9 {n}^{2} }{ {(n - m)}^{2} } = \frac{9 {m}^{2} }{ {(m - n)}^{2} } - \frac{9 {n}^{2} }{ {(m - n)}^{2} } = \frac{9 {m}^{2} - 9 {n}^{2} }{ {(m - n)}^{2} } = \frac{9(m - n)(m + n)}{ {(m - n)}^{2} } = \frac{9m + 9n}{m - n} [/tex]