证明2x²-4xy+4y²+2x+2的值总是不小于1
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证明2x²-4xy+4y²+2x+2的值总是不小于1
证明2x²-4xy+4y²+2x+2的值总是不小于1
证明2x²-4xy+4y²+2x+2的值总是不小于1
2x^2-4xy+4y^2+2x+2
=x^2-4xy+4y^2+x^2+2x+1+1
=(x+2y)^2+(x+1)^2+1
因为(x+2y)^2>=0,(x+1)^2>=0
所以原式>=1
所以2x²-4xy+4y²+2x+2的值总是不小于1
2x²-4xy+4y²+2x+2
=(x²-4xy+4y²)+(x²+2x+1)+1
=(x-2y)²+(x+1)²+1>=1
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证明2x²-4xy+4y²+2x+2的值总是不小于1
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