求(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)的值,其中X,Y满足丨x-y+1丨+(x-5)²=0

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求(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)的值,其中X,Y满足丨x-y+1丨+(x-5)²=0求(-3x²-4y)-(2x

求(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)的值,其中X,Y满足丨x-y+1丨+(x-5)²=0
求(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)的值,其中X,Y满足丨x-y+1丨+(x-5)²=0

求(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)的值,其中X,Y满足丨x-y+1丨+(x-5)²=0
丨x-y+1丨+(x-5)²=0
x-5=0
x=5
x-y+1=0
y=x+1=5+1=6
(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)
=-3x²-4y-2x²+5y-6+x²-5y-1
=(x²-3x²-2x²)+(5y-5y-4y)-(6+1)
=-4x²-4y-7
=-4x5²-4x6-7
=-100-24-7
=-131

(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)
=-3x²-4y-2x²+5y-6+x²-5y-1
=-4x²-4y-7

x-y+1=0
x-5=0
x=5
y=6

-4x²-4y-1
=-4×25-4×6-7

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(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)
=-3x²-4y-2x²+5y-6+x²-5y-1
=-4x²-4y-7

x-y+1=0
x-5=0
x=5
y=6

-4x²-4y-1
=-4×25-4×6-7
=-131

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|x-y+1|+(x-5)²=0
因为:|x-y+1|≥0,(x-5)²≥0
所以:x-y+1=0,x-5=0
x=5,y=6
(-3x²-4y)-(2x²-5y+6)+(x²-5y-1)
=-3x²-4y-2x²+5y-6+x²-5y-1
=x²-3x²-2x²+5y-4y-5y-1-6
=-4x²-4y-7
=-4×5²-4×6-7
=-131