第4章 4.6

(1)

4x2−4x+1=04x^2-4x+1=0

(2x−1)2=0x=12(2x-1)^2=0 \quad x=\dfrac{1}{2}

(2)

9x2+6x+1=09x^2+6x+1=0

(3x+1)2=0x=−13(3x+1)^2=0 \quad x=-\dfrac{1}{3}

(3)

50x2−20x+2=050x^2-20x+2=0

(5x−1)(10x−2)=0x=15(5x-1)(10x-2)=0 \quad x=\dfrac{1}{5}

(4)

12x2+27−36x=012x^2+27-36x=0

3(4x2−12x+9)=3(2x−3)2=0x=323(4x^2-12x+9)=3(2x-3)^2=0 \quad x=\dfrac{3}{2}

解説: r31bn1z