Условие:
1. При каких значениях переменной имеет смысл выражение 3/(y+7)?
2. Сократите дробь:
1) (15x^7 y^5)/(55x^4 y^6 )
2) (18ab-6b)/6ab
3) (a^2-1)/(3a+3)
4) (x^2-16x+64)/(64-x^2 )
3. Выполните вычитание:
1) (a-5)/(5a^3 )-(1-a)/a^4
2) 9/a-18/(a^2+2a)
3) x^2/(x^2-49)-x/(x+7)
4) 7b-(21b^2)/(3b+4)
4. Упростите выражение:
1) (a-18)/(2a-12)-(a-6)/(2a+12)+50/(a^2-36 )
2) (6c^3+3c)/(c^3-1)-(3c^2)/(c^2+c+1)
5. Известно, что (m+3n)/n=2. Найдите значение выражения:
1) m/n;
2) (m-5n)/m.
6. Постройте график функции y=(2x^2+5x)/x-(x^2-9)/(x+3).
Решение:
\[\boxed{\mathbf{1}\mathbf{.}\mathbf{\ }}\]
\[\frac{3}{y + 7}\]
\[y + 7 \neq 0\]
\[y \neq - 7\]
\[Выражение\ имеет\ смысл\ при\ \]
\[любом\ y,кроме\ y = - 7.\]
\[\boxed{\mathbf{2}\mathbf{.}\mathbf{\ }}\]
\[1)\frac{15x^{7}y^{5}}{55x^{4}y^{6}} = \frac{3x^{3}}{11y}\]
\[2)\frac{18ab - 6b}{6ab} = \frac{6b(3a - 1)}{6ab} =\]
\[= \frac{3a - 1}{a}\]
\[3)\frac{a^{2} - 1}{3a + 3} = \frac{(a - 1)(a + 1)}{3(a + 1)} =\]
\[= \frac{a - 1}{3}\]
\[4)\frac{x^{2} - 16x + 64}{64 - x^{2}} =\]
\[= \frac{(8 - x)^{2}}{(8 - x)(8 + x)} = \frac{8 - x}{8 + x}\]
\[\boxed{\mathbf{3}\mathbf{.}\mathbf{\ }}\]
\[1)\frac{a - 5^{\backslash a}}{5a^{3}} - \frac{1 - a^{\backslash 5}}{a^{4}} =\]
\[= \frac{a^{2} - 5a - 5 + 5a}{5a^{4}} = \frac{a^{2} - 5}{5a^{4}}\]
\[2)\frac{9}{a} - \frac{18}{a^{2} + 2a} =\]
\[= \frac{9^{\backslash a + 2}}{a} - \frac{18}{a(a + 2)} =\]
\[= \frac{9a + 18 - 18}{a(a + 2)} = \frac{9a}{a(a + 2)} =\]
\[= \frac{9}{a + 2}\]
\[3)\frac{x^{2}}{x^{2} - 49} - \frac{x^{\backslash x - 7}}{x + 7} =\]
\[= \frac{x^{2} - x^{2} + 7x}{(x - 7)(x + 7)} = \frac{7x}{x^{2} - 49}\]
\[4)\ 7b^{\backslash 3b + 4} - \frac{21b^{2}}{3b + 4} =\]
\[= \frac{21b^{2} + 28b - 21b^{2}}{3b + 4} = \frac{28b}{3b + 4}\]
\[\boxed{\mathbf{4}\mathbf{.}\mathbf{\ }}\]
\[1)\frac{a - 18}{2a - 12} - \frac{a - 6}{2a + 12} + \frac{50}{a^{2} - 36} =\]
\[= \frac{- 44}{2(a - 6)(a + 6)} =\]
\[= - \frac{22}{a^{2} - 36} = \frac{22}{36 - a²}\]
\[2)\frac{6c^{3} + 3c}{c^{3} - 1} - \frac{3c^{2}}{c^{2} + c + 1} =\]
\[= \frac{6c^{3} + 3c}{(c - 1)\left( c^{2} + c + 1 \right)} - \frac{3{c^{2}}^{\backslash c - 1}}{c^{2} + c + 1} =\]
\[= \frac{6c^{3} + 3c - 3c^{3} + 3c^{2}}{(c - 1)\left( c^{2} + c + 1 \right)} =\]
\[= \frac{3c^{3} + 3c^{2} + 3c}{(c - 1)\left( c^{2} + c + 1 \right)} =\]
\[= \frac{3c(c^{2} + c + 1)}{(c - 1)\left( c^{2} + c + 1 \right)} = \frac{3c}{c - 1}\]
\[\boxed{\mathbf{5}\mathbf{.}\mathbf{\ }}\]
\[\frac{m + 3n}{n} = 2\]
\[\frac{m}{n} + \frac{3n}{n} = 2\]
\[\frac{m}{n} + 3 = 2\]
\[\frac{m}{n} = - 1.\]
\[1)\frac{m}{n} = - 1.\]
\[2)\frac{m - 5n}{m} = \frac{m}{m} - \frac{5n}{m} = 1 - 5 \cdot \frac{n}{m} =\]
\[= 1 - 5 \cdot ( - 1) = 1 + 5 = 6.\]
\[\boxed{\mathbf{6}\mathbf{.}\mathbf{\ }}\]
\[y = \frac{2x^{2} + 5x}{x} - \frac{x^{2} - 9}{x - 3} =\]
\[= \frac{x(2x + 5)}{x} - \frac{(x - 3)(x + 3)}{x - 3} =\]
\[= 2x + 5 - (x + 3) =\]
\[= 2x + 5 - x - 3 = x + 2\]
\[y = x + 2;\ \ \ x \neq 0;\ \ x \neq 3\]