已知(x+3)的平方+|y+6|=0,求2(3x-3y)-4[3(x-y)-2(x-2y)]的值.
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![已知(x+3)的平方+|y+6|=0,求2(3x-3y)-4[3(x-y)-2(x-2y)]的值.](/uploads/image/z/1064155-67-5.jpg?t=%E5%B7%B2%E7%9F%A5%28x%2B3%29%E7%9A%84%E5%B9%B3%E6%96%B9%2B%7Cy%2B6%7C%3D0%2C%E6%B1%822%283x-3y%29-4%5B3%28x-y%29-2%28x-2y%29%5D%E7%9A%84%E5%80%BC.)
已知(x+3)的平方+|y+6|=0,求2(3x-3y)-4[3(x-y)-2(x-2y)]的值.
已知(x+3)的平方+|y+6|=0,求2(3x-3y)-4[3(x-y)-2(x-2y)]的值.
已知(x+3)的平方+|y+6|=0,求2(3x-3y)-4[3(x-y)-2(x-2y)]的值.
由题意,得
x+3=0,y+6=0
x=-3,y=-6
所以
原式=6x-6y-12(x-y)+8(x-2y)
=6x-6y-12x+12y+8x-16y
=2x-10y
=2×(-3)-10×(-6)
=-6+60
=54