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Resposta:

Letra "d"

Explicação passo-a-passo:

Elevando ambos os membros da equação

(A^B + B^A)^2 = 5^2, temos:

(A^B)^2 + 2(A^B)•(B^A) + (B^A)^2 = 25

Nota: (A^B)•(B^A)

A^2B + 2•11 + B^2A = 25

A^2B + B^2A = 25 - 22

A^2B + B^2A = 3

[(A^2B) + (B^2A) + 78]^(25%)

[(A^2B) + (B^2A) + 78]^(25/100)

[(A^2B) + (B^2A) + 78]^(1/4)

Nota: A^2B + B^2A = 3

[3 + 78]^(1/4)

81^(1/4)

[3^4]^1/4

3^(4/4)

3

Portanto, letra "d".

[tex]\displaystyle \sf A^{B}+B^{A} = 5 \ ;\ A^{B}\cdot B^{A} = 11 \ ;\ \ A,B\ \neq 0\\\\\\ \left[A^{2B}+B^{2A}+78\right]^{25\%}\\\\\ Fa{\c camos} :\\\\ A^{B} = x \ \ ;\ \ B^{A} = y \\\\ da{\'i}}: \\\\ x+y = 5 \\\\ (x+y)^2=5^2 \\\\ x^2+y^2+2xy = 25 \\\\ A^{2B}+B^{2A}+2\cdot \underbrace{\sf A^{B}\cdot B^{A}}_{11}=25 \\\\\\ A^{2B}+B^{2A} +2\cdot 11= 25 \\\\\ A^{2B}+B^{2A} = 25-22 = 3 \\\\ Da{\'i}}: \\\\ \left[\underbrace{\sf A^{2B}+B^{2A}}_{3}+78\right]^{25\%}[/tex]

[tex]\displaystyle \sf \boxed{\sf obs:\ 25\% = \frac{25}{100} = \frac{1}{4}}\\\\\\\ \left[3+78\right]^{\frac{1}{4}} = (81)^{\frac{1}{4}} \\\\\ (3^4)^{\frac{1}{4}} = 3^{\frac{4\cdot 1}{4}} = 3 \\\\\ Portanto : \\\\ \Large\boxed{\sf \ \left[A^{2B}+B^{2A}+78\right]^{25\%} = 3 \ }\checkmark[/tex]

letra d