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x²-(2k+3)x+k²+3k+2=0这方程怎么解?
人气:231 ℃ 时间:2020-03-29 18:51:02
解答
x²-(2k+3)x+k²+3k+2=0
→x²-[(k+1)+(k+2)]x+(k+1)(k+2)=0
→[x-(k+1)][x-(k+2)]=0
∴x1=k+1, x2=k+2.
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