Question

решите пожалуйста,хочу себя проверить!!!

Answer 1

 

$left { {{log_{5-x}frac{x+4}{(x-5)^{10}}geq-10,} atop {<var>x^3+8x^2+frac{50x^2-x-7}{x-7}leq1;</var>}} right.$

$begin{cases} 5-x>0,\5-xneq1,\ frac{x 4}{(x-5)^{10}}>0, \(x-5)^{10}neq0, \ x-7neq0; end{cases} begin{cases} x<5,\xneq4,\ x>-4, \xneq5, \ xneq7; end{cases} D=(-4;4)cup(4;5), \ </var><img src=$

$<var>log_{5-x}frac{x+4}{(x-5)^{10}}geq-10, \ left { {{0<5-x<1,} atop {frac{x+4}{(x-5)^{10}}leq(5-x)^{-10}};} right. left { {{4<x" title="</var></p> <p>[tex]<var>log_{5-x}frac{x+4}{(x-5)^{10}}geq-10, \ left { {{0<5-x<1,} atop {frac{x+4}{(x-5)^{10}}leq(5-x)^{-10}};} right. left { {{4<x" alt="</var></p> <p>[tex]<var>log_{5-x}frac{x+4}{(x-5)^{10}}geq-10, \ left { {{0<5-x<1,} atop {frac{x+4}{(x-5)^{10}}leq(5-x)^{-10}};} right. left { {{4<x" /></var></p> <p>[tex]<var>left { {{5-x>1,} atop {frac{x+4}{(x-5)^{10}}geq(5-x)^{-10}};} right. left { {{x<4,} atop {frac{x+4}{(x-5)^{10}}geqfrac{1}{(x-5)^{10}};} right. left { {{x<4,} atop {x+4geq1;} right. left { {{x<4,} atop {xgeq-3;} right. xin[-3;4);$

$begin{cases} 5-x>0,\5-xneq1,\ frac{x+4}{(x-5)^{10}}>0, \(x-5)^{10}neq0, \ x-7neq0; end{cases} begin{cases} x<5,\xneq4,\ x>-4, \xneq5, \ xneq7; end{cases} D=(-4;4)cup(4;5), \ </var>[tex]</p> <p>[tex]<var>log_{5-x}frac{x+4}{(x-5)^{10}}geq-10, \ left { {{0<5-x<1,} atop {frac{x+4}{(x-5)^{10}}leq(5-x)^{-10}};} right. left { {{4<x$

$<var>left { {{5-x>1,} atop {frac{x+4}{(x-5)^{10}}geq(5-x)^{-10}};} right. left { {{x<4,} atop {frac{x+4}{(x-5)^{10}}geqfrac{1}{(x-5)^{10}};} right. left { {{x<4,} atop {x+4geq1;} right. left { {{x<4,} atop {xgeq-3;} right. xin[-3;4);" /></var></p> <p>[tex]x^3+8x^2+frac{50x^2-x-7}{x-7}leq1, \ x^3+8x^2-1+frac{50x^2-x-7}{x-7}leq0, \ frac{(x^3+8x^2-1)(x-7)+50x^2-x-7}{x-7}leq0, \ frac{x^4+x^3-6x^2}{x-7}leq0, \ frac{x^2(x^2+x-6)}{x-7}leq0, \ x^2(x^2+x-6)(x-7)leq0, \ x^2(x^2+x-6)(x-7)=0, \ x_1=0, \ x^2+x-6=0, x_2=-3, x_3=2, \ x-7=0, x_4=7, \ (x+3)x^2(x-2)(x-7)leq0, \ xin(-infty;-3]cup[2;7)cup{0};$

 

$xin[2;4)cup{0}.$