Đáp án:
x2+2x+2|x+1|−2=0x2+2x+2|x+1|−2=0⇒(x2+2x+1)+2|x+1|−3=0⇒(x2+2x+1)+2|x+1|−3=0⇒(x+1)2+2|x+1|=3⇒(x+1)2+2|x+1|=3⇒|x+1|2+2|x+1|=3⇒|x+1|2+2|x+1|=3Đặt: |x+1|=t≥0|x+1|=t≥0 ta có: t2+2t=3t2+2t=3⇒t2+2t+1=4⇒t2+2t+1=4⇒(t+1)2=4⇔[t+1=2t+1=−2⇔[t=1t=−3(l)⇒(t+1)2=4⇔[t+1=2t+1=−2⇔[t=1t=−3(l)Vì: |x+1|=1⇔[x+1=1x+1=−1⇔[x=0x=−2|x+1|=1⇔[x+1=1x+1=−1⇔[x=0x=−2Vậy S={0;−2}
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