Question

решить системы логарифмических уравнений и неравенств.помогите пожалуйста ​

Answer 1

Ответ: A)

$\left \{ {lg x-lg y =7 { } \atop {lg x +lg y =5}} \right. \\\\\left \{ {lg x-lg y =7 { } \atop {lg x = 5-lg y}} \right. \\\\5- lg y -lg y=7\\y=\frac{1}{10} \\lgx=5-lg(\frac{1}{10} )\\ x=10^{6}\\(x,y)=( 10^{6},\frac{1}{10}) \\ \\\left \{ {lg 10^{6}-lg \frac{1}{10} =7 { } \atop {lg 10^{6}+lg \frac{1}{10} =5}} \right. \\\\\left \{ {{7=7} \atop {5=5}} \right. \\(x,y)=( 10^{6},\frac{1}{10})$

Б) $\left \{ {{log_{3}x+log_{3}y=2} \atop {x^{2} y-2y+9=0}} \right. \\\\\left \{ {{log_{3}xy=2 } \atop {x^{2} y-2y+9=0}} \right. \\\\\left \{ {{xy=3^{2} } \atop {x^{2} y-2y+9=0}} \right. \\\\\left \{{{xy=9 } \atop {x^{2} y-2y+9=0}} \right. \\\\(x1,y1)=(-2,-\frac{9}{2}) \\(x2,y2)=(1,9)\\\left \{ {{logx_{3} (-2)+{logx_{3}(-\frac{9}{2})=2 } \atop {(-2)^{2} *(-\frac{9}{2})-2* (-\frac{9}{2})+9=0}} \right. \\\\\left \{ {{{logx_{3}1+{logx_{3}9=2} \atop {1^{2}*9-2*9+9=0 }} \right.$$\left \{ {{logx_{3}(-2)+logx_{3} (-\frac{9}{2} )=2 } \atop {0=0}} \right. \\\\\left \{ {{2=2} \atop {0=0}} \right. \\\\(x,y)=(1,9)$

2.$\left \{ {{logx_{7} (6x-9)<{logx_{7} (2x-3) } \atop {logx_{\frac{1}{5} } (2-x)$≥$logx_{\frac{1}{5}(2x+4) } }} \right.$

Решаем систему неравенств

 $\left \{ {{x<\frac{3}{2} } \atop {x}} \right.$≥$-\frac{2}{3}$

Находи пересечение x ∈ [  $-\frac{2}{3} ,\frac{3}{2}$ 〉 , x ∈  〈 $\frac{3}{2} ,2$〉

x ∈ ∅