(3) (x+y+z)(-x+y+z)(x−y+z)(x+y−z)
=
=
{(y+z)²-x²}{x²-(y-z)²}
= {-x²+(y+z)²}{x²-(y-z)²}
{x+y+z)}{−x+y+z)}×{x−(y−z)}{x+(y-2)})
=−x++{(y+z)² + (y−2)²}x²−(y+z)²(y-2)²
=-x²+2(y²+z²) x²-(y²-2²) ²
=-x-y-z' +2xy'+2yz2+2z'x
(4) (x+y+1)(x²-xy+y²-x-y+1)
=
{x+(y+1)}{x²-(y+1)x+(y²-y+1)}
=x²+{(y+1)-(y+1)}x²+{(y²-y+1)-(y+1)²}x
+(y+1)(y²-y+1)
=x³+(-3y)x+y³+1
=x³+y²-3xy+1
数学 Ⅰ-5
1-
←平方の差の利用。
←(-x²+) (x²-■)の
形。
←(y+z)²(y-z)²
= {(y+z)(y-2)}²
←xについて整理し,
(x+)(x²-x+)
とみて展開。