已知Ix-2I+(y+1)²=0,你能求出代数式5x²+3xy-5y²-2xy+4y²-4x²的值吗?”
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 03:01:01
![已知Ix-2I+(y+1)²=0,你能求出代数式5x²+3xy-5y²-2xy+4y²-4x²的值吗?”](/uploads/image/z/4533018-42-8.jpg?t=%E5%B7%B2%E7%9F%A5Ix-2I%2B%28y%2B1%29%26%23178%3B%3D0%2C%E4%BD%A0%E8%83%BD%E6%B1%82%E5%87%BA%E4%BB%A3%E6%95%B0%E5%BC%8F5x%26%23178%3B%2B3xy-5y%26%23178%3B-2xy%2B4y%26%23178%3B-4x%26%23178%3B%E7%9A%84%E5%80%BC%E5%90%97%3F%E2%80%9D)
已知Ix-2I+(y+1)²=0,你能求出代数式5x²+3xy-5y²-2xy+4y²-4x²的值吗?”
已知Ix-2I+(y+1)²=0,你能求出代数式5x²+3xy-5y²-2xy+4y²-4x²的值吗?”
已知Ix-2I+(y+1)²=0,你能求出代数式5x²+3xy-5y²-2xy+4y²-4x²的值吗?”
Ix-2I+(y+1)²=0,绝对值与平方均为非负数,则有x=2,y=-1
5x²+3xy-5y²-2xy+4y²-4x²
=x^2-y^2+xy
=4-1-2
=1
由Ix-2I+(y+1)²=0得x=2;y=-1
所以5x²+3xy-5y²-2xy+4y²-4x²=x²+xy-y²=2^2+2*(-1)-(-1)^2=4-2-1=1