用因式分解法解下列因式：① 9（x+2）²=16（2x-5）²②（x+3）&#1

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① 9（x+2）²=16（2x-5）²
9（x+2）²-16（2x-5）²＝0
[3(x+2)+4(2x-5)][3(x+2)-4(2x-5)]=0
∴3(x+2)+4(2x-5)=0或3(x+2)-4(2x-5)=0
解得：x1=14/11,x2=26/5
②（x+3）²-5（x+3）+6=0
[(x+3)-2][(x+3)-3]=0
(x+3)-2=0或(x+3)-3=0
∴x1=-1,x2=0
======以下答案可供参考======
供参考答案1:
      3（x+2)=   4(2x-5)                  3(x+2)= -  4(  2x-5)
           3x+6=8x-20                          3X+6=-8X+20
                 5X=26                                 11X=14
                    X=26/5                                  X=14/11
   【  （x+3)-2]   [    (x+3)  -3  ]  =0
            (x+1)  *x=0
          x+1=0         x=0
           x=-1       Or    x=0