Ответ(ы) на вопрос:
Гость
1.
.........=(5x³(3x-5))/5x^5=(3x-5)/x²
........=((3y-1)(3y+1)) / (3(3y-1))=(3y+1)/3
........=((a+2)(a+2)) / ((2-a)(2+a))=(a+2)/(2-a)
2.
[latex] \frac{x+2}{x+3} - \frac{x-1}{x}= \frac{x(x+2)-(x-1)(x+3)}{x(x+3)}= \frac{ x^{2} +2x- x^{2} -3x+x+3}{x(x+3)}= \frac{3}{x(x+3)} [/latex]
[latex]2y- \frac{4 y^{2} }{2y-1} -1= \frac{2y(2y-1)-4y^{2}-(2y-1) }{2y-1}= \frac{4y^{2}-2y-4 y^{2} -2y+1}{2y-1}= \frac{1-4y}{2y-1} [/latex]
[latex] \frac{5a^{2} }{5ab-b^{2} } - \frac{b}{25a-5b}= \frac{5a^{2} }{b(5a-b)}- \frac{b}{5(5a-b)}= \frac{25 a^{2}- b^{2} }{5b(5a-b)}= \frac{(5a-b)(5a+b)}{5b(5a-b)}= \frac{5a+b}{5b} \\ \\ \frac{ x^{2} }{ x^{3}-x} + \frac{1}{2-2x}= \frac{ x^{2} }{x( x^{2} -1)} } + \frac{1}{2(1-x)}= \frac{ x^{2} }{x(x-1)(x+1)} - \frac{1}{x-1}= \\ \frac{ x^{2}-x(x+1) }{x(x-1)(x+1)}= \frac{ x^{2}- x^{2} -1 }{x(x-1)(x+1)}= -\frac{1}{x( x^{2} -1)} [/latex]
3.
[latex] .....=\frac{4a+1}{(a-2)(a^{2}+2a+4) }+ \frac{a}{(a^{2} +2a+4)}- \frac{1}{a-2} = \frac{4a+1+a(a-2)-( a^{2}+2a+4) }{( a-2)( a^{2}+2a+4) }= \\ \frac{4a+1+a^{2}-2a- a^{2}-2a-4 }{( a-2)( a^{2}+2a+4)} =- \frac{3}{ a^{3}-8 } [/latex]
Не нашли ответ?
Похожие вопросы