√2x√3 who can give the answer
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Answer:
〖(2+√3)〗^(x^2-2x+1)+〖(2-√3)〗^(x^2-2x-1)=4/(2-√3)
The above equation can be rewritten as
〖(2+√3)〗^(x^2-2x)*(2+√3)+〖(2-√3)〗^(x^2-2x)*(2-√3)-1 = 4/(2-√3)
Multiply the equation with 2-√3
We get〖(2+√3)〗^(x^2-2x)*(2+√3)(2-√3)+〖(2-√3)〗^(x^2-2x)*(2-√3)-1(2-√3)=4
Or 〖(2+√3)〗^(x^2-2x)*(4-3)+ 〖(2-√3)〗^(x^2-2x)*(2-√3)-1+1=4 ( Using (a+b)*(a-b)=a^2-b^2 in the first part on left side and law of indices in second part a^m*a^n=a^(m+n) and remembering that a^0=1)
〖(2+√3)〗^(x^2-2x)+〖(2-√3)〗^(x^2-2x)=4
In second part on left side we can multiply numerator and denominator with conjugate 2+√3
The equation now becomes
〖(2+√
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Step-by-step explanation:
root 2 × root 3 = root 6