Question

Sa se rezolve sistemul:
$\left \{ {{u+v=\sqrt{2} } \atop {u^{4}+v^{4}=4} \right.$

Answer 1

Explicație pas cu pas:

(u²)²+(v²)²=4⇒(u²+v²)²-2u²v²=4⇒((u+v)²-2uv)²-2u²v²=4

u+v=√2 |²⇒(u+v)²=2

⇒(4-2uv)²-2u²v²=4 , notam: uv=a , (u+v)=b

(4-2a)²-2a²=4⇒4a²-16a+16-2a²=4⇒2a²-16a+12=0 |:2

⇒a²-8a+6=0⇒(a-4-√10)(a-4+√10)=0

⇒b=±2

avem: u+v=√2 si uv=4+√10

⇒u=√2-v⇒v(√2-v)=4+√10⇒v²-v√2+4+√10=0⇒v∉R

u+v=-√2 si uv=4-√10

⇒v(v+√2)=√10-4⇒v²+v√2+4-√10=0

⇒v1=(-√2+(√(4√10-14)) )/2

⇒v2=(-√2-√(4√10-14)) )/2

u1=√2-v1⇒u1=(√2-(√(4√10-14))/2)

u2=√2-v2⇒u2=(√2+(√(4√10-14))/2)

Bafta!

Answer 2

Răspuns:

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Explicație pas cu pas: