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当IX-2I+(Y+1/2)^2=0时,5(X^2-3Y^2)-[-(X^2-2XY-Y^2)+(5X^2-2XY=4Y^)]求值.
人气:260 ℃ 时间:2020-10-01 14:49:37
解答
IX-2I+(Y+1/2)^2=0,则X-2=0,X=2,Y+1/2=0,y= -1/2
5(X^2-3Y^2)-[-(X^2-2XY-Y^2)+(5X^2-2XY-4Y^)]
=5X^2-15Y^2-(-X^2+2XY+Y^2+5X^2-2XY-4Y^)
=5X^2-15Y^2+X^2-2XY-Y^2-5X^2+2XY+4Y^
=x^2-12y^2
=2^2-12*(-1/2)^2
=4-3
=1
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