Question

даю 100 баллов алгебра​

Answer 1

$\displaystyle\bf\\4.1\\\\ODZ: \ 3x+4 > 0 \ \ \ \Rightarrow \ \ \ x > -1\frac{1}{3} \\\\\log_{\frac{1}{3} } (3x+4)=-2\\\\3x+4=\Big(\frac{1}{3} \Big)^{-2} \\\\3x+4=9\\\\\\3x=5\\\\\boxed{x=1\frac{2}{3}} \\\\4.2\\\\ODZ: \ \left \{ {{x > 0} \atop {x-3 > 0}} \right. \ \ \ \Rightarrow \ \ \ x > 3\\\\\\\log_{2} x+\log_{2} (x-3)=2\\\\\log_{2} \Big[x\cdot(x-3)\Big]=2\\\\x\cdot(x-3)=4\\\\x^{2} -3x-4=0\\\\Teorema \ Vieta :\\\\x_{1} =4 \ \ \ , \ \ \ x_{2} =-1- \ neyd\\\\Otvet \ : \ 4$

$\displaystyle\bf\\4.3\\\\ODZ: \ \left \{ {{3x+4 > 0} \atop {x^{2} -4x-14 > 0}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x > -1\dfrac{1}{3} } \atop {x^{2} -4x-14 > 0}} \right.\\\\\\\log_{\frac{1}{3} } (3x+4)=\log_{\frac{1}{3} } (x^{2} -4x-14)\\\\x^{2} -4x-14=3x+4\\\\x^{2} -4x-14-3x-4=0\\\\x^{2} -7x-18=0\\\\Teorema \ Vieta \ :\\\\x_{1}=9 \ \ \ ; \ \ \ x_{2} =-2- \ neyd\\\\Otvet \ : \ 9\\\\\\5.1$

$\displaystyle\bf\\ODZ: \ \left \{ {{x-4 > 0} \atop {8-x > 0}} \right. \ \ \ \left \{ {{x > 4} \atop {x < 8}} \right. \\\\\\\log_{0,9} (x-4)\geq \log_{0,9}(8-x)\\\\0 < 0,9 < 1 \ \ \ \Rightarrow \ \ \ x-4\leq 8-x\\\\x+x\leq 8+4\\\\2x\leq 12\\\\x\leq 6\\\\Otvet \ : \ x\in\Big(4 \ ; \ 6\Big]\\\\\\5.2\\\\ODZ: \ x > 0\\\\\lg^{2} x-lgx-6 > 0$

$\displaystyle\bf\\(\lg x-3)\cdot (\lg x+2) > 0\\\\\\+ + + + + (-2) - - - - - (3) + + + + + \\\\\\\left \{ {{\lg x > -2} \atop {\lg x < 3}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x > 0,01} \atop {x < 1000}} \right. \\\\\\Otvet \ : \ x\in(0,01 \ ; \ 1000)$