Risolvere: $\sqrt{10+\sqrt{32+\sqrt{13+\sqrt{x}}}}=4$
Esercizio
$\sqrt{10+\sqrt{32+\sqrt{13+\sqrt{x}}}=4}$
Soluzione passo-passo
Impara online a risolvere i problemi di passo dopo passo. (10+(32+(13+x^(1/2))^(1/2))^(1/2))^(1/2)=4. Applicare la formula: x^a=b\to \left(x^a\right)^{inverse\left(a\right)}=b^{inverse\left(a\right)}, dove a=\frac{1}{2}, b=4, x^a=b=\sqrt{10+\sqrt{32+\sqrt{13+\sqrt{x}}}}=4, x=10+\sqrt{32+\sqrt{13+\sqrt{x}}} e x^a=\sqrt{10+\sqrt{32+\sqrt{13+\sqrt{x}}}}. Applicare la formula: x+a=b\to x=b-a, dove a=10, b=16, x+a=b=10+\sqrt{32+\sqrt{13+\sqrt{x}}}=16, x=\sqrt{32+\sqrt{13+\sqrt{x}}} e x+a=10+\sqrt{32+\sqrt{13+\sqrt{x}}}. Applicare la formula: a+b=a+b, dove a=16, b=-10 e a+b=16-10. Applicare la formula: x^a=b\to \left(x^a\right)^{inverse\left(a\right)}=b^{inverse\left(a\right)}, dove a=\frac{1}{2}, b=6, x^a=b=\sqrt{32+\sqrt{13+\sqrt{x}}}=6, x=32+\sqrt{13+\sqrt{x}} e x^a=\sqrt{32+\sqrt{13+\sqrt{x}}}.
(10+(32+(13+x^(1/2))^(1/2))^(1/2))^(1/2)=4
Risposta finale al problema
$x=9$