x(x+1)+(x+2)+(x+3)+(x+4)+(x+5) = 27求x

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x(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=27求xx(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=27求xx(x+1)+(x+2)+(x+3)+(x+4)+(x+5)

x(x+1)+(x+2)+(x+3)+(x+4)+(x+5) = 27求x
x(x+1)+(x+2)+(x+3)+(x+4)+(x+5) = 27求x

x(x+1)+(x+2)+(x+3)+(x+4)+(x+5) = 27求x
x²＋x＋x＋2＋x＋3＋x＋4＋x＋5=27
x²＋5x＋14=27
x²＋5x－13=0
x=(－5±√77)/2

无解

X=2

没有实数解吧！
化简得：X²+5X－13=0
X=－b±﹙b²﹣4ac﹚½／2a=﹣5±(77)½／2

x(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=27
x^2+x+x+2+x+3+x+4+x+5=27
x^2+5x+14=27
x^2+5x-13=0
解b^2-4ac 即 5^2-...

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x(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=27
x^2+x+x+2+x+3+x+4+x+5=27
x^2+5x+14=27
x^2+5x-13=0
解b^2-4ac 即 5^2-4*1*(-13)=77>0,则方程有两个不相同的实数根，
所以， x={-b±√（b^2－4ac）}/2a
={-5±√（5^2-4*1*（-13））}/2a
x1=(√77-5)/2 x2=-(√77+5)/2

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