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(4π-πr^2)=πr^2×4π怎么得到(r^2+2r-4)(r^2-2r-4)=0?
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■按你的算式
(4π-πr²)=πr²×4π
两边同除以π,
4-r²=4πr²
整理得
(4π+1)r²=4
■你把算是写错了,应该是
(4π-πr²)²=πr²×4π,
π²(4-r²)²=4π²r²
两边同除以π²,
(4-r²)²=4r²
整理得
(4-r²)²
-4r²=0,
(4-r²+2r)(4-r²-2r)=0,
(-4+r²-2r)(-4+r²+2r)=0,
(r²-2r-4)(r²+2r-4)=0
(4π-πr²)=πr²×4π
两边同除以π,
4-r²=4πr²
整理得
(4π+1)r²=4
■你把算是写错了,应该是
(4π-πr²)²=πr²×4π,
π²(4-r²)²=4π²r²
两边同除以π²,
(4-r²)²=4r²
整理得
(4-r²)²
-4r²=0,
(4-r²+2r)(4-r²-2r)=0,
(-4+r²-2r)(-4+r²+2r)=0,
(r²-2r-4)(r²+2r-4)=0
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