Question

解不等式(x+1)²(x-1)(x+2)(x-3)≤0

急急急，过程

Answer 1

穿针法，奇穿偶不穿
所以(x+1)²这里的x=-1不穿
其他零点是-2,1,3
是小于等于0
所以x≤-2，1≤x≤3

Answer 2

121x^4-4y^2
=(11x^2+2y)(11x^2-2y)

4/25a^2-X^2y^2
=(2/5a+Xy)(2/5a-Xy)

(m+n)^2-n^2
=(m+n+n)(m+n-n)
=m(m+2n)

4(2p+3q)^2-(3p-q)^2
=(4p+6q+3p-q)(4p+6q-3p+q)
=(7p+5q)(p+7q)

81a^4-2y^2
不能分解

8y^4-2y^2
=y^2(81y^2-2)

3ax^2-3ay^4
=2a(x+y^2)(x-y^2)

m^4-1
=(m^2+1)(m^2-1)
=(m^2+1)(m+1)(m-1)

12x^2-3y^2
=3(4x^2-y^2)
=3(2x+y)(2x-y)

(2x+y)^2-(x+2y)^2
=(2x+y+x+2y)(2x+y-x-2y)
=3(x+y)(x-y)