Nhớ....................Vote....................Nha
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Dk: x≥−1=>x+1≥0 pt⟺5(√1+x3−2x−2)=4x4−25x3+18x2−10x−15⟺5(1+x3−4x2−8x−4)√1+x3+2x+2=(x2−5x−3)(4x2−5x+5)
⟺5(x+1)(x2−5x−3)√1+x3+2x+2=(x2−5x−3)(4x2−5x+5)
x2−5x−3=0.v.5(x+1)√1+x3+2x+2=4x2−5x+5
Th1:x2−5x−3=0=>.....
Ta có: 4x2−5x+5=(2x−54)2+5516>3
Th2:5(x+1)√1+x3+2x+2=4x2−5x+5>3=>5(x+1)>3√1+x3+6x+6
⟺0>3√1+x3+x+1(Vo.li)=>Vo.nghiem
Vậy x=5+√372,x=5−√372