a)
1+√3 = x+2
x = 1+√3-2
x = √3-1
=======
b)
1/(√3+x) = √3/x I*x(√3+x) Z:x ≠ 0
√3(√3+x) = x
3+√3x = x
√3x-x = -3
x(√3-1) = -3 /:(√3-1)
x = -3/(√3-1) * (√3+1)/(√3+1) = -3(√3+1)/(3-1) = -3(√3+1)/2 = -1,5(√3+1)
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a)
1+√3 = x+2
x = 1+√3-2
x = √3-1
=======
b)
1/(√3+x) = √3/x I*x(√3+x) Z:x ≠ 0
√3(√3+x) = x
3+√3x = x
√3x-x = -3
x(√3-1) = -3 /:(√3-1)
x = -3/(√3-1) * (√3+1)/(√3+1) = -3(√3+1)/(3-1) = -3(√3+1)/2 = -1,5(√3+1)