先化简,再求值[(x+y)(x-2y)-(x+2y)^2]/2分之1y,其中x=-1,y=4分之1
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解
原式
=[(x²-xy-2y²)-(x²+4xy+4y²)]/(1/2y)
=(-5xy-6y²)/(1/2y)
=(-5x-6y)×2
=-10x-12y
=-10×(-1)-12×(1/4)
=10-3
=7
10.3×9.7
=(10+0.3)+(10-0.3)
=10²-0.3²
=100-0.09
=99.91
(2m+n-p)(2m-n+p)
=[2m+(n-p)][2m-(n-p)]
=4m²-(n-p)²
=4m²-n²+2pn-p²
原式
=[(x²-xy-2y²)-(x²+4xy+4y²)]/(1/2y)
=(-5xy-6y²)/(1/2y)
=(-5x-6y)×2
=-10x-12y
=-10×(-1)-12×(1/4)
=10-3
=7
10.3×9.7
=(10+0.3)+(10-0.3)
=10²-0.3²
=100-0.09
=99.91
(2m+n-p)(2m-n+p)
=[2m+(n-p)][2m-(n-p)]
=4m²-(n-p)²
=4m²-n²+2pn-p²
追问
10.3*9.7那题错了,=(10+0.3)(10-0.3),答案没错啊
追答
嗯嗯
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