Verificăm dacă lungimile laturilor ar fi un triplet pitagoreic de forma
(3k, 4k, 5k), k∈ℕ*.
[tex]\it x+1=3k \Longrightarrow k = \dfrac{x+1}{3} \ \ \ \ (*)
\\\;\\ \\\;\\
x+4=4k \Longrightarrow k = \dfrac{x+4}{4}\ \ \ \ (**)[/tex]
[tex]\it (*),\ (**) \Longrightarrow \dfrac{x+1}{3} = \dfrac{x+4}{4} \Longrightarrow 4x+4=3x+12 \Longrightarrow
\\\;\\ \\\;\\
\Longrightarrow 4x-3x = 12-4 \Longrightarrow x=8[/tex]
x+1 = 8+1 = 9
x+4 = 8+4 = 12
x+7 = 8+7 = 15
Tripletul (9, 12, 15) este pitagoreic.
Deci, soluția problemei este x = 8.
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[tex]\it E = (sin^2 30^o +cos^2 60^o) - (sin20^o-cos70^o)[/tex]
[tex]\it cos70^o = sin(90^o-70^o) =sin20^o[/tex]
Acum, expresia devine:
[tex]\it E = (sin^2 30^o +cos^2 60^o) - (sin20^o- sin20^o) = \dfrac{1}{4} + \dfrac{1}{4} -0 =\dfrac{1}{2}[/tex]
