展开 (2xy+y^2)|(1,0)|

解答

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(2 x y + y^{2}) ∣ (1, 0) ∣

2 ∣ (1 \cdot 0,) ∣ x y + ∣ (1 \cdot 0,) ∣ y^{2}

求解步骤

(2 x y + y^{2}) ∣ (1, 0) ∣

= ∣ (1 \cdot 0,) ∣ (2 x y + y^{2})

使用分配律:

a (b + c) = ab + a c

a = ∣ (1, 0) ∣, b = 2 x y, c = y^{2}

= ∣ (1, 0) ∣ \cdot 2 x y + ∣ (1, 0) ∣ y^{2}

= 2 ∣ (1 \cdot 0,) ∣ x y + ∣ (1 \cdot 0,) ∣ y^{2}

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

e x p an d π (x)^{2} (3 x + 1) - 90 π

e x p an d \frac{( 4 t ^{2} + 45 ) ^{\frac{3}{2}} - 343}{12}

e x p an d \frac{( x + 2 ) ( x + 3 ) ( x + 4 )}{( x + 1 )}

e x p an d (\frac{3}{2}, \frac{1}{4}) \cdot (x - \frac{π}{4}, y)