Let frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B cd | vedclass

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(D) Let $x = 2^{1/4}$. Then the expression becomes $\frac{7}{x^2 + x + 1}$.Multiplying numerator and denominator by $(x - 1)$,we get $\frac{7(x - 1)}{

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(D) Let $x = 2^{1/4}$. Then the expression becomes $\frac{7}{x^2 + x + 1}$.Multiplying numerator and denominator by $(x - 1)$,we get $\frac{7(x - 1)}{x^3 - 1}$.Since $x = 2^{1/4}$,$x^3 = 2^{3/4}$.So,$\frac{7(2^{1/4} - 1)}{2^{3/4} - 1} = A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4}$.Multiplying by $(2^{3/4} + 1)$,we get $7(2^{1/4} - 1)(2^{3/4} + 1) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2^{3/4} - 1)(2^{3/4} + 1)$.$7(2^{1} + 2^{1/4} - 2^{3/4} - 1) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2^{3/2} - 1)$.$7(1 + 2^{1/4} - 2^{3/4}) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2 \cdot 2^{1/2} - 1)$.Expanding and equating coefficients,we find $A = 1, B = -3, C = 2, D = 1$.Thus,$A + B + C + D = 1 - 3 + 2 + 1 = 1$.

Let $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4}$,then find the value of $A + B + C + D$.

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