1 фотка
1. а, б, г, е, ж
2. а) любое значение, кроме 0; б) тоже самое что и в а; в) любое число, кроме 3; г) любое число, кроме 2; д) и е) любое число
2 фотка (там немного размыто, так что проверь условие моих примеров)
1 задание
a) ![\frac{6a^{4}b }{12a^{2}b^{3} } =\frac{a^{2} }{2b^{2} } \frac{6a^{4}b }{12a^{2}b^{3} } =\frac{a^{2} }{2b^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B6a%5E%7B4%7Db%20%7D%7B12a%5E%7B2%7Db%5E%7B3%7D%20%20%7D%20%3D%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2b%5E%7B2%7D%20%7D)
б) ![\frac{a^{3}+3a^{2}b}{a^{2}-9b^{2}} = \frac{a^{2}(a+3b) }{(a+3b)(a-3b)} = \frac{a^{2}}{a-3b} \frac{a^{3}+3a^{2}b}{a^{2}-9b^{2}} = \frac{a^{2}(a+3b) }{(a+3b)(a-3b)} = \frac{a^{2}}{a-3b}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E%7B3%7D%2B3a%5E%7B2%7Db%7D%7Ba%5E%7B2%7D-9b%5E%7B2%7D%7D%20%3D%20%5Cfrac%7Ba%5E%7B2%7D%28a%2B3b%29%20%7D%7B%28a%2B3b%29%28a-3b%29%7D%20%20%3D%20%5Cfrac%7Ba%5E%7B2%7D%7D%7Ba-3b%7D)
в) ![\frac{x^{2}+4x+4 }{2x^{2}-8 } = \frac{(x+2)^{2} }{2(x^{2} -4)} =\frac{(x+2)^{2} }{2(x-2)(x+2)} = \frac{x+2}{2(x-2)} \frac{x^{2}+4x+4 }{2x^{2}-8 } = \frac{(x+2)^{2} }{2(x^{2} -4)} =\frac{(x+2)^{2} }{2(x-2)(x+2)} = \frac{x+2}{2(x-2)}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D%2B4x%2B4%20%7D%7B2x%5E%7B2%7D-8%20%7D%20%3D%20%5Cfrac%7B%28x%2B2%29%5E%7B2%7D%20%7D%7B2%28x%5E%7B2%7D%20-4%29%7D%20%3D%5Cfrac%7B%28x%2B2%29%5E%7B2%7D%20%7D%7B2%28x-2%29%28x%2B2%29%7D%20%3D%20%5Cfrac%7Bx%2B2%7D%7B2%28x-2%29%7D)
2 задание
а) ![\frac{2a+5}{a^{2} -9} -\frac{a+2}{a^{2}-9 } =\frac{2a+5-a-2}{a^{2}-9 } =\frac{a+3}{(a-3)(a+3)} = \frac{1}{a-3} \frac{2a+5}{a^{2} -9} -\frac{a+2}{a^{2}-9 } =\frac{2a+5-a-2}{a^{2}-9 } =\frac{a+3}{(a-3)(a+3)} = \frac{1}{a-3}](https://tex.z-dn.net/?f=%5Cfrac%7B2a%2B5%7D%7Ba%5E%7B2%7D%20-9%7D%20-%5Cfrac%7Ba%2B2%7D%7Ba%5E%7B2%7D-9%20%7D%20%3D%5Cfrac%7B2a%2B5-a-2%7D%7Ba%5E%7B2%7D-9%20%7D%20%3D%5Cfrac%7Ba%2B3%7D%7B%28a-3%29%28a%2B3%29%7D%20%3D%20%5Cfrac%7B1%7D%7Ba-3%7D)
б) ![\frac{c^{2}+8c }{4-c^{2} } +\frac{4-4c}{4-c^{2} } =\frac{c^{2}+8c+4-4c}{4-c^{2} } =\frac{c^{2}+4c+4}{(2-c)(2+c)} = \frac{(c+2)^{2}}{(2-c)(2+c)} = \frac{c+2}{2-c} \frac{c^{2}+8c }{4-c^{2} } +\frac{4-4c}{4-c^{2} } =\frac{c^{2}+8c+4-4c}{4-c^{2} } =\frac{c^{2}+4c+4}{(2-c)(2+c)} = \frac{(c+2)^{2}}{(2-c)(2+c)} = \frac{c+2}{2-c}](https://tex.z-dn.net/?f=%5Cfrac%7Bc%5E%7B2%7D%2B8c%20%7D%7B4-c%5E%7B2%7D%20%7D%20%2B%5Cfrac%7B4-4c%7D%7B4-c%5E%7B2%7D%20%7D%20%3D%5Cfrac%7Bc%5E%7B2%7D%2B8c%2B4-4c%7D%7B4-c%5E%7B2%7D%20%7D%20%3D%5Cfrac%7Bc%5E%7B2%7D%2B4c%2B4%7D%7B%282-c%29%282%2Bc%29%7D%20%3D%20%5Cfrac%7B%28c%2B2%29%5E%7B2%7D%7D%7B%282-c%29%282%2Bc%29%7D%20%20%3D%20%5Cfrac%7Bc%2B2%7D%7B2-c%7D)