Nhớ....................Vote....................NhaDk: x≥−1=>x+1≥0pt⟺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+5Th1:x2−5x−3=0=>.....Ta có: 4x2−5x+5=(2x−54)2+5516>3Th2:5(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.nghiemVậy x=5+√372,x=5−√372
Nhớ....................Vote....................Nha Dk:
x≥−1=>x+1≥0pt⟺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+5Th1:x2−5x−3=0=>.....Ta có:
4x2−5x+5=(2x−54)2+5516>3Th2:5(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.nghiemVậy
x=5+√372,x=5−√372