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a/ (x+4)(x+5)(x+7)(x+8)=4
⇔(x+4)(x+8)(x+5)(x+7)=4
⇔(x2+12x+32)(x2+12x+35)=4
Đặt x2+12x+32=t,t≥−4 được phương trình:
t(t+3)−4=0⇔t2+3t−4=0⇔t=1,t=−4
Với t=1:x2+12x+32=1⇔x2+12x+31=0⇔x=−6−√5,x=−6+√5
Với t=−4:x2+12x+32=−4⇔x2+12x+36=0⇔(x+6)2=0⇔x=−6
b/ (x−2)2+(x−3)4=1
Đặt t=x+−3−22=x−52⇒x=t+52
Thay ẩn t:
(t+52−2)4+(t+52−3)4=1
⇔(t+12)4+(t−12)4=1
⇔[(t+12)2−(t−12)2]2+2(t+12)2.(t−12)2=1
4t2+3t2−78=0⇔t2=14⇔t=≠12t=12⇒x=3,t=−12⇒x=2
c/ (x2−4x+5)2−(x−1)(x−3)=4
⇔(x2−4x+5)2−(x2−4x+5)−2=0
Đặt (x2−4x+5)=t,t≥1
Có phương trình trung gian: t2−t−2=0⇔t=−1(L),t=2
(x2−4x+5)=2⇔x2−4x+3=0⇔x=1,x=3
d/ (x2+3x−4)2+3(x2+3x−4)=x+4
⇔(x2+3x−4)2+4(x2+3x−4)=x2+4x
⇔[(x2+3x−4)2–x2]+4[(x2+3x−4)–x]=0
⇔[(x2+3x−4)–x][x2+3x−4+x+4]=0
⇔(x2+3x−4)(x2+4x)=0
⇔x1=−1−√5,x2=−1+√5,x3=0,x4=−4
e/ 6x4−35x3+62x2−35x+6=0
⇔6x2−35x+62−35.1x+6x2=0
⇔6(x2+1x2)−35(x+1x)+62=0
đặt x+1x=t thì |t|≥2⇒x2+1x2=t2−2
được phương trình trung gian 6(t2−2)−35t+62=0⇔6t2−35t+50=0
t1=35−512=52,t2=35+512=103
x+1x=52⇔x2−52x+1=0⇔x=2,x=12
x+1x=103⇔x2−103x+1=0⇔x=3,x=13
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