展开 (2e^{2t}cos(t)-sin(t)*e^{2t})^2

解答

展开

(2 e^{2 t} cos (t) - sin (t) \cdot e^{2 t})^{2}

4 e^{4 t} cos^{2} (t) - 4 e^{4 t} cos (t) sin (t) + e^{4 t} sin^{2} (t)

求解步骤

(2 e^{2 t} cos (t) - sin (t) e^{2 t})^{2}

= (2 e^{2 t} cos (t) - e^{2 t} sin (t))^{2}

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 2 e^{2 t} cos (t), b = sin (t) e^{2 t}

= (2 e^{2 t} cos (t))^{2} - 2 \cdot 2 e^{2 t} cos (t) sin (t) e^{2 t} + (sin (t) e^{2 t})^{2}

化简

(2 e^{2 t} cos (t))^{2} - 2 \cdot 2 e^{2 t} cos (t) sin (t) e^{2 t} + (sin (t) e^{2 t})^{2} : 4 e^{4 t} cos^{2} (t) - 4 e^{4 t} cos (t) sin (t) + e^{4 t} sin^{2} (t)

= 4 e^{4 t} cos^{2} (t) - 4 e^{4 t} cos (t) sin (t) + e^{4 t} sin^{2} (t)

e x p an d 2 ce n t ro (\frac{1}{2}, - 3)

e x p an d (\frac{yy - 2 y + 1}{4})^{2}

\frac{d}{d y} ((x (y)) (\frac{y ^{6}}{6} + \frac{1}{16 y ^{4}})^{2})

e x p an d \frac{2 ( 2 i z ^{2} - 15 z ^{2} + 15 i z + 16 z - 14 i )}{( z - i ) ^{5}}

e x p an d (\frac{2 x + 1}{4 x}) \cdot (\frac{- 16}{x ( 1 + x ) ( x - 4 ) ^{2}})