展开 r(r+2cos(θ)=0)

解答

展开

r (r + 2 cos (θ) = 0)

r^{2} + 2 r cos (θ) = 0

求解步骤

r (r + 2 cos (θ) = 0)

= r (r + 2 \cdot 0 = cos (θ))

使用分配律:

a (b + c) = ab + a c

a = r, b = r, c = 2 cos (θ) = 0

rr + r \cdot 2 cos (θ) = 0

rr + 2 r cos (θ) = 0

乘开

rr + 2 r cos (θ) : r^{2} + 2 r cos (θ)

r^{2} + 2 r cos (θ) = 0

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

e x p an d (e u + e + 1)^{2}

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

e x p an d (A x + B) e^{4 x}

e x p an d \frac{4 ( x ^{0.5} - 3 ) ^{2}}{x}