Question

Решите прошу 1,2
Заранее спасибо

Answer 1

1.

$\ttdisplaystyle\frac{t^{2}-8t }{4} =0\\ \\ t^{2} -8t=0\\ \\ t(t-8)=0\\ \\ t_{1}=0\\  \\ t-8=0\\ \\ x_{2} =8\\ \\ .\\ \frac{2x-3}{4} =\frac{3x-1}{3} \\ \\ 3(2x-3)=4(3x-1)\\ \\ 6x-9=12x-4\\ \\ 12x-6x=4-9\\ \\ 6x=-5\\ \\ x=-\frac{5}{6} \\ \\ .\\ \frac{x+1}{x-2} -\frac{x}{x+2} =\frac{12}{x^{2}-4 } \\ \\\frac{x+1}{x-2} -\frac{x}{x+2} = \frac{12}{(x-2)(x+2)} \\ \\  \frac{(x+1)(x+2)-x(x-2)}{(x-2)(x+2)}= \frac{12}{(x-2)(x+2)}\\ \\ \frac{x^{2}+2x+x+2-x^{2}+2x-12}{(x+2)(x-2)} =0\\ \\[tex]\frac{5x-10}{(x+2)(x-2)} =0\\ \\ \frac{5(x-2)}{(x+2)(x-2)} =0\\ \\ \frac{5}{x-2} =0\;\;\;\;\;\;\;\;\;\;\;\;x\neq 0\\ \\ x-2\neq 0\\ \\ x\neq 2$

x  (-∞; 2)∪(2; ∞)

......

3.

$5^{-2} =\frac{1}{5^{2} }=\frac{1}{25} \\ \\ 12*3^{-3}=\frac{3*4}{3^{3} }  =\frac{4}{3^{2} } =\frac{4}{9} \\ \\ (27*3^{-2} )^{-2}=(3^{3} *3^{-2} )^{-2}=(3^{3-2})^{-2}=3^{-2}=\frac{1}{3^{2} } =\frac{1}{9} \\ \\ 2^{-3} *2^{2} =2^{-3+2}=2^{-1}=\frac{1}{2}$