Question

24. Fie a = √(1-√2)² + √(√2-√3)² + √(√3-2)². Demonstrați că a € N.​

Answer 1

Răspuns:

a = 1 ∈ N

Explicație pas cu pas:

a = √(1-√2)² + √(√2-√3)² + √(√3-2)²

a = I1-√2I + I√2-√3I + I√3-2I

√2 > 1 =>  I1-√2I = √2-1

√3 > √2 => I√2-√3I = √3-√2

√3 < 2 =>  I√3-2I = 2-√3 =>

a = √2-1 + √3-√2 + 2-√3

a = 2-1 => a = 1 ∈ N

Answer 2

$\it 1 < 2 \Rightarrow \sqrt1 < \sqrt2 \Rightarrow 1 < \sqrt2 \Rightarrow 1-\sqrt2 < 0 \Rightarrow |1-\sqrt2|=-1+\sqrt2\\ \\ \sqrt{(\sqrt2-\sqrt3)^2}=|\underbrace{\it\sqrt2-\sqrt3}_{ < 0}|=-\sqrt2+\sqrt3\\ \\ \sqrt{(\sqrt3-2)^2}=|\underbrace{\it \sqrt3-2}_{ < 0}|=-\sqrt3+2\\ \\ \\ a=-1+\sqrt2-\sqrt2+\sqrt3-\sqrt3+2=1\in\mathbb{N}$