x2+2x+1x2+2x+2+x2+2x+2x2+2x+3=76a)x2+2x+1x2+2x+2+x2+2x+2x2+2x+3=76
Ta có:x2+2x+2=x2+2x+1+1=(x+1)2+1≥1(a≥1)x2+2x+2=x2+2x+1+1=(x+1)2+1≥1(a≥1)
Đặt a=x2+2x+2(a≥1)a=x2+2x+2(a≥1)
pt⇔a−1a+aa+1=76pt⇔a-1a+aa+1=76
⇔6(a−1)(a+1)+6a2=7a(a+1)⇔6(a-1)(a+1)+6a2=7a(a+1)
⇔6(a2−1)+6a2=7a2+7a⇔6(a2-1)+6a2=7a2+7a
⇔12a2−6=7a2+7a⇔12a2-6=7a2+7a
⇔5a2−7a−6=0⇔5a2-7a-6=0
⇔5a2−10a+3a−6=0⇔5a2-10a+3a-6=0
⇔5a(a−2)+3(a−2)=0⇔5a(a-2)+3(a-2)=0
⇔(a−2)(5a+3)=0⇔(a-2)(5a+3)=0
Vì a≥1a≥1
⇒5a+3≥8>0⇒5a+3≥8>0
⇔a−2=0⇔a-2=0
⇔a=2(tmđk)⇔a=2(tmđk)
⇔x2+2x+2=2⇔x2+2x+2=2
⇔x2+2x=0⇔x2+2x=0
⇔x(x+2)=0⇔x(x+2)=0
⇔[x=0x+2=0⇔[x=0x+2=0
⇔[x=0x=−2⇔[x=0x=-2
Vậy phương trình có tập nghiệm S={0;−2}.
HOI LAG TÍ NHA :((
