展开 (e^{tcos(x)}(t)-e^{tsin(x)}(t))^2

解答

展开

(e^{t c o s (x)} (t) - e^{t s i n (x)} (t))^{2}

e^{2 t c o s (x)} t^{2} - 2 e^{t c o s (x) + t s i n (x)} t^{2} + e^{2 t s i n (x)} t^{2}

求解步骤

(e^{t c o s (x)} (t) - e^{t s i n (x)} (t))^{2}

去除括号:

(a) = a

= (e^{t c o s (x)} t - e^{t s i n (x)} t)^{2}

使用完全平方公式:

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

a = e^{t c o s (x)} t, b = e^{t s i n (x)} t

= (e^{t c o s (x)} t)^{2} - 2 e^{t c o s (x)} t e^{t s i n (x)} t + (e^{t s i n (x)} t)^{2}

化简

(e^{t c o s (x)} t)^{2} - 2 e^{t c o s (x)} t e^{t s i n (x)} t + (e^{t s i n (x)} t)^{2} : e^{2 t c o s (x)} t^{2} - 2 e^{t c o s (x) + t s i n (x)} t^{2} + e^{2 t s i n (x)} t^{2}

= e^{2 t c o s (x)} t^{2} - 2 e^{t c o s (x) + t s i n (x)} t^{2} + e^{2 t s i n (x)} t^{2}

e x p an d 2 ((3, - 4, 5, 0, 0) - 5 (4, 8, - 2, - 1, 3) + (- 4, 7, - 5))

e x p an d \frac{sec ( x )}{2 sin ( x )}

e x p an d \frac{( x + 3 ) ^{2}}{x - 5}

e x p an d \frac{9 ( x + 2 ) ^{3} ( x - 6 )}{2 ( x - 7 ) ^{\frac{1}{2}}}

e x p an d \frac{x - a}{( x ^{2} - 2 x ) ( x + b )}