Question

please solve to get the brainliest​

Answer 1

$(2+\sqrt3)^{(x^2-2x+1)}+(2-\sqrt3)^{(x^2-2x-1)}=\frac{2}{2-\sqrt3}</p><p>\\</p><p>(2+\sqrt3)^{(x^2-2x+1)}\times (2-\sqrt3)+(2-\sqrt3)^{(x^2-2x-1)}\times {(2-\sqrt3)}=2</p><p>\\\bold{a^m.a^n=a^{m+n}}\\</p><p>(2+\sqrt3)^{(x^2-2x)}.(2+\sqrt3).(2-\sqrt3)+(2-\sqrt3)^{(x^2-2x)}.(2-\sqrt3)^{-1}.(2-\sqrt3)=2\\</p><p>\bold{(a+b).(a-b)=a^2-b^2}</p><p>\\</p><p>\bold{a^{-1}.a=1}</p><p>\\</p><p>(2+\sqrt3)^{(x^2-2x)}.(4-3)+(2-\sqrt3)^{(x^2-2x)}=2===> \bold{E\:01}</p><p>\\</p><p>Let\:\bold{n}=x^2-2x\\a=(2+\sqrt3)\\b=(2-\sqrt3)\\</p><p>\bold{a^n+b^n=2}==> \bold{E\:02}\\</p><p>By\:observation\:we\:make\:a\:relationship\:between\:\bold{a}\:and\:\bold{b}</p><p>\\</p><p>i.e\:</p><p>\\</p><p>a.b=1</p><p>\\</p><p>(2+\sqrt3)(2-\sqrt3)=1</p><p>\\</p><p>a=\frac{1}{b}\\from\:E\:02\:we\:get</p><p>\\</p><p>(\frac{1}{b})^n+b^n=2</p><p>\\</p><p>\bold{Let\:b^n=t} \\ \frac{1}{t}+t=2\\t^2-2t+1=0\\t^2-t-t+1=0\\t(t-1)-1(t-1)=0\\(t-1)(t-1)=0\\(t-1)^2=0\\t=1\\</p><p>which\:is\:\\t=b^n=1\\ \bold{=>b^n=(2+\sqrt3)^{x^2-2x}=1}</p><p>\\Apply\:Logarithm\:on\:both\:sides\\log((2+\sqrt3)^{x^2-2x})=log\:1\\(x^2-2x).log(2+\sqrt3)=0\\log(2+\sqrt3)\:never\:be\:zero\\So,\:\:\:\bold{x^2-2x}\:\:\:must\:be\:zero\\\\ \bold{x^2-2x=0}\\\bold{x(x-2)=0}\\\bold{x=0\:or\:2}$

The real values of x are 0 & 2.

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