(y+2)(y-2)--(y-1)(y+5)123的2次方减去122乘124(2x +y+z)(2x-y-z)(x+1)(x+3)-(x-2)的2次方 +(x+2)(x-2) (a+2b-1)(-2+ab)(2+ab)
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(y+2)(y-2)--(y-1)(y+5)123的2次方减去122乘124(2x +y+z)(2x-y-z)(x+1)(x+3)-(x-2)的2次方 +(x+2)(x-2) (a+2b-1)(-2+ab)(2+ab)
(y+2)(y-2)--(y-1)(y+5)
123的2次方减去122乘124
(2x +y+z)(2x-y-z)
(x+1)(x+3)-(x-2)的2次方 +(x+2)(x-2)
(a+2b-1)
(-2+ab)(2+ab)
(y+2)(y-2)--(y-1)(y+5)123的2次方减去122乘124(2x +y+z)(2x-y-z)(x+1)(x+3)-(x-2)的2次方 +(x+2)(x-2) (a+2b-1)(-2+ab)(2+ab)
解(1)(y+2)(y-2)--(y-1)(y+5)
=y²-4 -y²-4y+5 = 1-4y
(2)123的2次方减去122乘124
=123²-(123-1)(123+1) = 123² -123²+1 = 1
(3)2x +y+z)(2x-y-z)
=4x²-(y+z)² = 4x²-y²-z²-2yz
(4)(x+1)(x+3)-(x-2)的2次方 +(x+2)(x-2)
=x²+ 4x +3 -x²+4x-4+x²-4 = x²+8x -5
(5﹚
(6)(-2+ab)(2+ab)
=a²b²- 4
(1)原式=y^2-4 -y^2-4y+5 = 1-4y
(2)原式=123^2-(123-1)(123+1) = 123^2 -123^2+1 = 1
(3)原式=4x^2-(y+z)^2 = 4x^2-y^2-z^2-2yz
(4)原式=x^2 + 4x +3 -x^2+4x-4+x^2-4 = x^2 +8x -5
(5)原式=a+2b-1
(6)原式=a^2b^2 - 4
(1)原=y^2-4 -y^2-4y+5 = 1-4y
(2)原=123^2-(123-1)(123+1) = 123^2 -123^2+1 = 1
(3)原=4x^2-(y+z)^2 = 4x^2-y^2-z^2-2yz
(4)原=x^2 + 4x +3 -x^2+4x-4+x^2-4 = x^2 +8x -5
(5)原=a+2b-1
(6)原=a^2b^2 - 4