Question

1. x^{2} = 1 +3 +5 +...+ 101

2. 6(x-1)^{2} = (2 + \sqrt10 ) * \sqrt14 -2\sqrt40

Answer 1

1) Metoda 1. 
$1+3+5+...+101= \frac{102\cdot51}{2}=51^2$ ⇒ $x=51$
Metoda 2.
Avem formula $1+3+5+...+(2n-1)=n^2$
⇒ $2x-1=101 \\ x=51$
2) $6(x-1)^2=(2+ \sqrt{10})\cdot \sqrt{14}-2 \sqrt{40} \\ 6(x-1)^2=2 \sqrt{14}+\sqrt{140}-2 \sqrt{40} \\ 6(x-1)^2=2( \sqrt{14}+ \sqrt{35}- \sqrt{40}) \\ 3(x-1)^2=\sqrt{14}+ \sqrt{35}- \sqrt{40}$
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