Question

まい

赤く矢印引いたところの変形を詳しく教えてください🙏

D
= (x²+x)²-14(x²+x)+48
= (x²+x-6) (x²+x-8)
=(x-2)(x+3)(x²+x-8)
(2)xの2次式にする。
5x={(y+1)x+(y+1)} (yx+1)+yx
S
=y(y+1)x²+{(y+1)+y(y+1)+y}x+(y+1)
=y(y+1)x²+{(y+1)²+y}x+(y+1)
=
{(y+1)x+1}{yx+(y+b)}
= (xy+x+1)(xy+y+1)
(3) 5=6x²+(5y-1)x-(6y²-5y+1)
=6x²+(5y-1)x-(2y-1)(3y-1)
2X
3-
3y-1
9y-3
- (2y-1) → -4y+2
5y-1
→
=
{2x+(3y-1)}{3x-(2y-1)}
= (2x+3y-1)(3x-2y+1)+)
(4) 5=(x²+4)²-8x²+4x²=(x²+4)² − 4x²
= (x²+4)²-(2x)² = (x²+4+2x) (x²+4-2x)
2)
= {(x+y+z)-x} {(x+y-
x²} −(y+z)(y²—yz+z²
= (y+z)(x²+y²+z²+2x=
+xz+x²) − (y+z)(y²–
= (y+z) (3x² + y² +₂²+3.
-(y+z)(y²-yz+z²)
= (y+z)(3x²+3xy+3yz
=3(y+z){x²+(y+z)x-
=3(y+z)(x+y)(x+z)
=3(x+y)(y+z)(z+x)
2
√6-√3-√2+1
2
解き方 与式=-
11
(√2+1)(√3+1)-
(1+√2)+
(√2-1)(√3
(√2+1)(√2-1)(√/3
√6-√2-√3+1

Answer 1

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