Question

Vreau un răspuns de la cineva care se pricepe foarte bine la matematică, vă rog eu!!!(Punctele încercuite, E1 i,k E2;b,c și E4 b)(în special domnului AugustinDevian)

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$4^{x*x}=1\\4^{x*x}=4^0\\x^2=0\\x=0$

x^2=x*x

$3^{2x-1}=81\\3^{2x-1}=3^4\\2x-1=4\\2x=4+1=5\\x=\frac{5}{2}$

$5^{x*x+1}=3125\\5^{x*x+1}=5^5\\x^2+1=5\\x^2=5-1=4\\x^2=4\\x_1=\sqrt{4} =2\\x_2=-\sqrt{4} =-2$

$2^{x*x-8x+4,5}=16*\sqrt{2} \\2^{x*x-8x+4,5}=2^4\sqrt{2} \\2^{x*x-8x+4,5}=2^4*2^{\frac{1}{2} }\\2^{x*x-8x+4,5}=2^{\frac{9}{2} }\\x^2-8x+4,5=\frac{9}{2} \\x^2-8x+\frac{9}{2} =\frac{9}{2} \\x^2-8x=0\\x(x-8)=0\\x_1=0\\x_2=x-8=0\\x_2=8$

$3^{2x+1}+3^{2x}-3^{2x-1}=297\\(3^2+3-1)*3^{2x-1}=297\\(9+3-1)*3^{2x-1}=297\\11*3^{2x-1}=297 |:11\\3^{2x-1}=27\\3^{2x-1}=3^3\\2x-1=3\\2x=3+1=4\\2x=4\\x=\frac{4}{2} =2$