ta có pt⇔4x4−25x3+18x2−5−5√1+x3=0 ⇔4x4−25x3+18x2−10x−15−5(√1+x3−2(x+1))=0 ⇔(x2−5x−3)(4x2−5x+5−5(x+1)√1+x3+2(x+1))=0 ⇔x2−5x−3=0hoặc 4x2−5x+5=5(x+1)√1+x3+2(x+1) (1)VT(1) ≥5516>3VP= 52−5√1+x32(√1+x3+2(x+1)≤52⇒ (1) VNo$\Rightarrow x=\frac{5\pm \sqrt{37}}{2
ta có pt
⇔4x4−25x3+18x2−5−5√1+x3=0 ⇔4x4−25x3+18x2−10x−15−5(√1+x3−2(x+1))=0 ⇔(x2−5x−3)(4x2−5x+5−5(x+1)√1+x3+2(x+1))=0 ⇔x2−5x−3=0hoặc
4x2−5x+5=5(x+1)√1+x3+2(x+1) (1)VT(1)
≥5516>3VP=
52−5√1+x32(√1+x3+2(x+1)≤52⇒ (1) VNo$\Rightarrow x=\frac{5\pm \sqrt{37}}{2
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