thay m=-1 vào x2(2m-1)x-m=0 ta có:
x2+(-3)x+1=0ΔΔ=5
⇔\orbr⎧⎨⎩x=3+√52x=3−√52⇔\orbr{x=3+52x=3−52
b) A=x21+x22−x1x2=(x1+x2)2−2x1x2−x1x2=(x1+x2)2−3x1x2x12+x22−x1x2=(x1+x2)2−2x1x2−x1x2=(x1+x2)2−3x1x2
Vi-et \hept{x1+x2=1−2mx1x2=−m\hept{x1+x2=1−2mx1x2=−m
=> A=(1−2m)2−3(−m)=4m2−4m+1+3m=4m2−m+
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