$x+\sqrt{5+\sqrt{x-1} } = 6$
như vậy lên bậc 4 rồi –  Cảii Bắp 24-10-15 09:01 PM
thử chuyển vế bình phương xem –  dolaemon 24-10-15 08:51 PM
chị lớp 9 nà =)) –  Cảii Bắp 24-10-15 08:06 PM
ko,e mới lớp 8 thôi,ko cần cảm ơn đâu –  ๖ۣۜTQT☾♋☽ 24-10-15 07:46 PM
em cảm ơn ạ –  Cảii Bắp 24-10-15 07:35 PM
cho 2 cái $$ vào đầu và cuối công thức đi bạn –  ๖ۣۜTQT☾♋☽ 24-10-15 07:11 PM
chuyển vế bình phương lên bậc 4:
$x^4-24x^3+206x^2-754x+962=0\Leftrightarrow (x^2-11x+26)(x^2-13x+37)=0$
$\sqrt{5+\sqrt{x-1}}=6-x$ (Đk : $1 \leq x \leq 6$)
$\Leftrightarrow 5+\sqrt{x-1}=x^2-12x+36$
$\Leftrightarrow x^2-11x+26=\sqrt{x-1}+x-5$
$\Leftrightarrow x^2-11x+26=\frac{-x^2+11x-26}{\sqrt{x-1}+5-x}$
$\Leftrightarrow(x^2-11x+26)(1+\frac{1}{\sqrt{x-1}+5-x})=0$
giải ra chỉ có $\color{blue}{x= \frac{11-\sqrt{17}}{2}}$ thỏa mãn đk
Đáp án chính xác rồi bạn. Cảm ơn nhiều nhé !! –  Cảii Bắp 24-10-15 10:57 PM

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