化简a^4-2a^2c^2+a^2b^2-b^2c^2+c^4=0
因式分解:4x^2-4x-y^2+4y-3
人气:116 ℃ 时间:2020-03-28 10:02:12
解答
a^4-2a^2c^2+a^2b^2-b^2c^2+c^4=0=>[a^4-2a^2c^2+c^4]+a^2b^2-b^2c^2=0=>[a^2-c^2]^2+b^2[a^-c^2]=0=>[a^2-c^2][a^2-c^2+b^2]=0=>[a-c][a+c][a^2-c^2+b^2]=0-----------4x^2-4x-y^2+4y-3 =4x^2-4x+1-y^2+4y-4=[2x-1]...
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