化简(x^3+3x^2+x-1/x+1)+1/2[(x-1+√2）^2+(x-1-√2)^2]
2个回答
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（x^3+3x^2+x-1）/（x+1）+1/2[(x-1+√2）^2+(x-1-√2)^2]
＝(x+1)(x^2+2x-1)/(x+1)+1/2[2(x-1)^2+4]
=x^2+2x-1+(x-1)^2+2
=2x^2+2
＝(x+1)(x^2+2x-1)/(x+1)+1/2[2(x-1)^2+4]
=x^2+2x-1+(x-1)^2+2
=2x^2+2
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展开全部
(x³+3x²+x-1)/(x+1) + 1/2 [(x-1+√2)²+(x-1-√2)²]
= (x³+3x²+x-1)/(x+1) + 1/2 [(x-1+√2)²+(x-1-√2)²]
={x(x+1)²+(x+1)(x-1)}/(x+1) + 1/2(2x²+2)
= x(x+1) + (x-1) + (x²+1)
= 2x²+2x
= (x³+3x²+x-1)/(x+1) + 1/2 [(x-1+√2)²+(x-1-√2)²]
={x(x+1)²+(x+1)(x-1)}/(x+1) + 1/2(2x²+2)
= x(x+1) + (x-1) + (x²+1)
= 2x²+2x
追答
(x³+3x²+x-1)/(x+1) + 1/2 [(x-1+√2)²+(x-1-√2)²]
= (x³+3x²+x-1)/(x+1) + 1/2 [(x-1+√2)²+(x-1-√2)²]
={x(x+1)²+(x+1)(x-1)}/(x+1) + 1/2(x²-2x-3)
= x(x+1) + (x-1) + (1/2x²-x+3/2)
= 3/2x²+x+1/2
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