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Đk: x≥13Ta có: pt⟺9x2−9x−10x2−4x+7=x√3x−1−2x
⟺(3x−5)(3x+2)x2−4x+7=x(3x−5)√3x−1+2
⟺x=53.v.3x+2x2−4x+7=x√3x−1+2(1)
pt(1)⟺(3x+2)√3x−1+2(3x+2)=x(x2−4x+7)
⟺(3x+2)√3x−1=x3−4x2+x−4
⟺x3−4x2+x−4−(3x+2)√3x−1=0
⟺x3−4x2+x−4−(3x+2)(x−2)+(3x+2)(x−2−√3x−1)=0
⟺x3−7x2+5x+(3x+2)(x−2−√3x−1)=0
⟺x(x−2−√3x−1)(x−2+√3x−1)+(3x+2)(x−2−√3x−1)=0
⟺(x−2−√3x−1)[x2−2x+x√3x−1+3x+2]=0
⟺(x−2−√3x−1)[x2+x+2+x√3x−1]=0
Do x≥13 nên [...]>0
Suy ra x−2−√3x−1=0⟺(x−2)2=3x−1⟺x2−7x+5=0(x>=2)
⟺x=7+√292.v.x=7−√292(l)
Bạn tự KL nhé. Nhớ Vote! và chấm đúng nha!