Question

ммм 100 баллов, за эти 2 простых уравнения, с полным решением пожалуйста​

Answer 1

$2)3*4^{x} +\frac{1}{3}*9^{x+2}=6*4^{x+1}-\frac{1}{2}*9^{x+1}\\\\\frac{1}{3}*9^{x}*9^{2}+\frac{1}{2}*9^{x}*9=6*4^{x}*4-3*4^{x}\\\\27*9^{x}+4,5*9^{x}=24*4^{x}-3*4^{x}\\\\31,5*9^{x}=21*4^{x} |:4^{x}>0\\\\31,5*(\frac{9}{4})^{x}=21\\\\(\frac{9}{4})^{x}= \frac{2}{3}\\\\(\frac{3}{2})^{2x}=(\frac{3}{2})^{-1}\\\\2x=-1\\\\x=-0,5$

ОДЗ : 6 - x ≥ 0        - x ≥ - 6       x ≤ 6

$3)x^{2}*4^{\sqrt{6-x}}+4^{x+2}=2^{4+2\sqrt{6-x}} +x^{2}*4^{x}\\\\4^{x}*4^{2}-x^{2}* 4^{x}=2^{4} *2^{2\sqrt{6-x}}-x^{2}*4^{\sqrt{6-x}}\\\\16*4^{x}-x^{2}*4^{x}=16*4^{\sqrt{6-x}}-x^{2}*4^{\sqrt{6-x}}\\\\4^{x}(16-x^{2})=4^{\sqrt{6-x}}(16-x^{2})$

$(16-x^{2} )(4^{x}-4^{\sqrt{6-x}})=0\\\\\left[\begin{array}{ccc}16-x^{2}=0 \\4^{x}-4^{\sqrt{6-x}}=0\end{array}\right\\\\1)16-x^{2}=0\\\\(4-x)(4+x)=0\\\\x_{1} =4\\\\x_{2}=-4\\\\2)4^{x}-4^{\sqrt{6-x }}=0\\\\4^{\sqrt{6-x}}=4^{x}\\\\\sqrt{6-x}=x,x\geq0\\\\(\sqrt{6-x})^{2}=x^{2}\\\\6-x=x^{2}\\\\x^{2}+x-6=0\\\\x_{3}=-3-neyd\\\\x_{4}=2-teoremaVieta\\\\Otvet:\boxed{-4;2;4}$