1)
\\ x^2+5x+7=1 \\ x^2+5x+6=0 \\ x_1=-2 \ x_2=-3" alt="5\cdot 2^{x^2+5x+7}+2^{x^2+5x+9}-2^{x^2+5x+10}=2 \\ 5\cdot 2^{x^2+5x+7}+2^2\cdot 2^{x^2+5x+7}-2^3\cdot 2^{x^2+5x+7}=2 \\ 2^{x^2+5x+7}\cdot(5+2^2-2^3)=2 \\ 2^{x^2+5x+7}=2 \ => \\ x^2+5x+7=1 \\ x^2+5x+6=0 \\ x_1=-2 \ x_2=-3" align="absmiddle" class="latex-formula">
3)
\\ x+\frac{1}{6}=-2x+4 \\ 3x=\frac{23}{6} \\ x=\frac{23}{18}" alt="\sqrt[3]{0,2}\cdot \sqrt{0,2^{2x-\frac{1}{3}}}=\sqrt[3]{0,04^{-3x+6}} \\ 0,2^{\frac{1}{3}} \cdot 0,2^{x-\frac{1}{6}}=\sqrt[3]{0,2^{-6x+12}} \\ 0,2^{\frac{1}{3}} \cdot 0,2^{x-\frac{1}{6}}=0,2^{-2x+4} \\ 0,2^{x+\frac{1}{6}}=0,2^{-2x+4} \ => \\ x+\frac{1}{6}=-2x+4 \\ 3x=\frac{23}{6} \\ x=\frac{23}{18}" align="absmiddle" class="latex-formula">
4)
