$x(x-4)(x^{2}-4x+9)=6\sqrt{4-x} -6\sqrt{x} -4$
thì bạn cứ đăng lên đi..... –  Nguyễn Nhung 21-04-16 08:58 PM
nếu ra đc đúng kết quả thì dở đc tuốt –  nguyenthiquynhphuong 21-04-16 07:31 PM
bài này mình nhân liên hợp khá dở , bạn có cần xem ko ? –  tran85295 21-04-16 06:53 PM
umk..... –  Nguyễn Nhung 20-04-16 10:18 PM
3 người bạn nhau à? –  dolaemon 20-04-16 08:42 PM
hâm j đấy m –  Nguyễn Nhung 19-04-16 09:36 PM
như 2 con hâm thân nhau là pai, trừ t –  nguyenthiquynhphuong 19-04-16 09:31 PM
umk............ –  Nguyễn Nhung 19-04-16 08:59 PM
oh, chắc bạn thân à? –  dolaemon 19-04-16 08:28 PM
noi chung đều là nhung học cung 1 lop –  nguyenthiquynhphuong 19-04-16 07:30 PM
k p đâu nha bạn! –  Nguyễn Nhung 18-04-16 10:25 PM
cho hỏi chút: nhungangels có phải là nhungevil ko??? –  dolaemon 18-04-16 10:24 PM
(đk $x \in [0;2]$
$VT=(x^2-4x+3+2\sqrt5)(x^2-4x+6-2\sqrt 5)+2-6\sqrt 5$
$\Rightarrow(x^2-4x+3+2\sqrt5)(x^2-4x+6-2\sqrt 5)=6(\sqrt{4-x}-\sqrt x-1+\sqrt 5)$
$\Leftrightarrow \frac{(x^2-4x+3+2\sqrt5)(x^2-4x+6-2\sqrt 5)}{6}=(\sqrt{4-x}-\sqrt x)+(\sqrt 5-1)$(*)
$VP$(*)$=\frac{(\sqrt{4-x}-\sqrt x)^2-(\sqrt 5-1)^2}{(\sqrt{4-x}-\sqrt x)-(\sqrt 5-1)}$
$=\frac{-2\left[\sqrt{x(4-x)} +1-\sqrt 5 \right]}{(\sqrt{4-x}-\sqrt x)-(\sqrt 5-1)}$
$=\frac{2(x^2-4x+6-2\sqrt 5)}{\left[\sqrt{x(4-x)}-(1-\sqrt 5) \right] \left[(\sqrt{4-x}-\sqrt x)-(\sqrt 5-1) \right]}$ (đk mẫu khác 0)
Đặt $x^2-4x+6-2\sqrt 5$ làm nhân tử chung, dễ cm đống còn lại $>0$
$\Rightarrow x=2 \pm \sqrt{2(\sqrt 5-1)}$
Mà nghiệm bé ko thõa đk $\Rightarrow \boxed{x=2 + \sqrt{2(\sqrt 5-1)}}$
Siêu kinh điển –  nguyenthiquynhphuong 22-04-16 06:30 PM

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