第15章 15.1

(1)

x=3⋅9+1⋅13+1=7x = \dfrac{3 \cdot 9 + 1 \cdot 1}{3 + 1} = 7

(2)

x=3⋅9−7⋅13−7=−5x = \dfrac{3 \cdot 9 - 7 \cdot 1}{3 - 7} = -5

(3)

x=5⋅9−1⋅15−1=11x = \dfrac{5 \cdot 9 - 1 \cdot 1}{5 - 1} = 11

(4)

x=9+12=5x = \dfrac{9 + 1}{2} = 5

解説: r31bn1z