-(x²-2xy+2y²﹚+[﹣3xy-2﹙x²-y﹚],其中x=﹣1,y=2先简化再求值
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/16 10:15:23
-(x²-2xy+2y²﹚+[﹣3xy-2﹙x²-y﹚],其中x=﹣1,y=2先简化再求值
-(x²-2xy+2y²﹚+[﹣3xy-2﹙x²-y﹚],其中x=﹣1,y=2
先简化再求值
-(x²-2xy+2y²﹚+[﹣3xy-2﹙x²-y﹚],其中x=﹣1,y=2先简化再求值
原式=-x²+2xy-2y²﹣3xy-2x²+2y
=-3x²-xy-2y²+2y
=-3+2-8+4
=-5
原式=-x²+2xy-2y²-3xy-2x²+2y
=-3x²-xy-2y²+2y
代入=-5
-(x²-2xy+2y²﹚+[﹣3xy-2﹙x²-y﹚],其中x=﹣1,y=2
=-x²+2xy-2y²+[-3xy-2x²+2y]
=-x²+2xy-2y²-3xy-2x²+2y
=(-1-2)x²+(2-3)xy-2y²+2y
=-3x²+(-...
全部展开
-(x²-2xy+2y²﹚+[﹣3xy-2﹙x²-y﹚],其中x=﹣1,y=2
=-x²+2xy-2y²+[-3xy-2x²+2y]
=-x²+2xy-2y²-3xy-2x²+2y
=(-1-2)x²+(2-3)xy-2y²+2y
=-3x²+(-xy)-2y²+2y
=-3x²-xy-2y²+2y
-3x²-xy-2y²+2y
=-3*(-1)²-(-1)*2-2*2²+2*2
=-3-(-2)-8+4
=-1-8+4
=-9+4
=-5
收起