代数式x-(x-1)/2与(x+2)/5-2的值互为相反数,则x的值为多少
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![代数式x-(x-1)/2与(x+2)/5-2的值互为相反数,则x的值为多少](/uploads/image/z/2779695-63-5.jpg?t=%E4%BB%A3%E6%95%B0%E5%BC%8Fx-%28x-1%29%2F2%E4%B8%8E%28x%2B2%29%2F5-2%E7%9A%84%E5%80%BC%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2C%E5%88%99x%E7%9A%84%E5%80%BC%E4%B8%BA%E5%A4%9A%E5%B0%91)
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代数式x-(x-1)/2与(x+2)/5-2的值互为相反数,则x的值为多少
代数式x-(x-1)/2与(x+2)/5-2的值互为相反数,则x的值为多少
代数式x-(x-1)/2与(x+2)/5-2的值互为相反数,则x的值为多少
思路:代数式x-(x-1)/2与(x+2)/5-2的值互为相反数,
【x-(x-1)/2】+【(x+2)/5-2】=0,转化为一元一次方程问题
【x-(x-1)/2】+【(x+2)/5-2】=0
10x-5(x-1)+2(x+2)-20=0
10x-5x+5+2x+4-20=0
10x-5x+2x=20-5-4
7x=11
x=11/7