proszę o pomoc
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Zad.1. a)
3*4^{2+\sqrt2}:4\sqrt2^{\sqrt{32}}=\frac{3*4^2*4^{\sqrt2}}{4*2^{\frac{1}{2}*4\sqrt2}}=
=\frac{3*4*(2^2)^{\sqrt2}}{2^{2\sqrt2}}=\frac{12*2^{2\sqrt2}}{2^{2\sqrt2}}=12
Zad. 2 b)
-2log_{4x}6=\frac{1}{2}
log_{4x}(\frac{1}{36})=\frac{1}{2}
(4x)^{\frac{1}{2}}=\frac{1}{36}
(4x)^{\frac{1}{2}}=(\frac{1}{1296})^{\frac{1}{2}}
4x=\frac{1}{1296}/:4
x=\frac{1}{5184}
Zad.2 c)
4^4x+4^4x+4^4x+4^4x=2^{103}
4*4^4x=2^{103}
4^5x=2^{103}
2^{10}x=2^{103}/:2^{10}
x=2^{93}
Zad. 2 a)
13^{x^2-2x}=13^1
x^2-2x=1
x^2-2x-1=0
\Delta=4+4=8
\sqrt{\Delta}=2\sqrt2
x_1=\frac{2-2\sqrt2}{2}=1-\sqrt2
x_2=1+\sqrt2
Zad. 1 d)
\sqrt{4-2\sqrt3}+2\sqrt{7+4\sqrt3}=\sqrt{1-2\sqrt3+3}+2\sqrt{4+4\sqrt3+3}=
=\sqrt{(1-\sqrt3)^2}+2\sqrt{(2+\sqrt3)^2}=
=|1-\sqrt3|+2|2+\sqrt3|=-(1-\sqrt3)+2(2+\sqrt3)=
-1+\sqrt3+4+2\sqrt3=3+3\sqrt3