Ответ:
Объяснение:
|5x-12|=7+|6-5x|\\f(x)=|5x-12|\\Ox:|5x-12|=0=>x=2,6\\Oy:13\\g(x)=7+|6-5x|\\Oy:13\\" alt="1)|5x-13|-|6-5x|=7<=>|5x-12|=7+|6-5x|\\f(x)=|5x-12|\\Ox:|5x-12|=0=>x=2,6\\Oy:13\\g(x)=7+|6-5x|\\Oy:13\\" align="absmiddle" class="latex-formula">
Вершина g(x) берёт начало в точке 1,2 =>x≤1.2
x^2-4\sqrt{x^2} -1=0\\\sqrt{x^2}=t,t>0=>t^2-4t-1 =0=>t=2+\sqrt{5}=>\\ \sqrt{x^2} =2+\sqrt{5}=>x=-2-\sqrt{5};x=2+\sqrt{5}\\3)-2\frac{x}{|x|} -2x=x^2+2<=>-2\frac{x}{|x|}-2x-x^2=2<=>\\ <=>-2x-2x|x|-x^2|x|=2|x|\\-2x-2x^2-x^3=2x<=>x(x^2+2x+4)=0=>x=0\\-2x+2x^2+x^3=-2x<=>x^2(2+x)=0=>x=-2;x=0\\ODZ:x\neq 0=>x=-2\\4)5^{|4x-6|}=25^{3x-4}<=>|4x-6|=6x-8\\4x-6=6x-6<=>-2x=-2=>x=1;x\geq 1,5\\6-4x=6x-8<=>-10x=-14=>x=\frac{7}{5};x<1,5" alt="2)x^2-4|x|-1=0<=>x^2-4\sqrt{x^2} -1=0\\\sqrt{x^2}=t,t>0=>t^2-4t-1 =0=>t=2+\sqrt{5}=>\\ \sqrt{x^2} =2+\sqrt{5}=>x=-2-\sqrt{5};x=2+\sqrt{5}\\3)-2\frac{x}{|x|} -2x=x^2+2<=>-2\frac{x}{|x|}-2x-x^2=2<=>\\ <=>-2x-2x|x|-x^2|x|=2|x|\\-2x-2x^2-x^3=2x<=>x(x^2+2x+4)=0=>x=0\\-2x+2x^2+x^3=-2x<=>x^2(2+x)=0=>x=-2;x=0\\ODZ:x\neq 0=>x=-2\\4)5^{|4x-6|}=25^{3x-4}<=>|4x-6|=6x-8\\4x-6=6x-6<=>-2x=-2=>x=1;x\geq 1,5\\6-4x=6x-8<=>-10x=-14=>x=\frac{7}{5};x<1,5" align="absmiddle" class="latex-formula">