Một câu thôi $\sqrt[3]{162x^3+2}-\sqrt{27x^2-9x+1}=1$
link đây bạn http://123doc.org/document/1203772-phuong-phap-dung-luong-lien-hop-de-giai-phuong-trinh-vo-ti.htm  :)
Phương trình được viết lại:
$\sqrt[3]{162x^3+2}-2=\sqrt{27x^2-9x+1}-1$
Thẫy rõ: $(\sqrt[3]{162x^3+2})^{2}+2\sqrt[3]{162x^3+2}+4>0$ (do với mọi a,b ta luôn có $a^2+b^2+ab\geq 0$ ở đây dấu bằng không xảy ra)
và $\sqrt{27x^2-9x+1}+1>0$
nên phương trình có thể viết lại thành
$\frac{(\sqrt[3]{162x^3+2}-2)[(\sqrt[3]{162x^3+2})^2+2\sqrt[3]{162x^3+2}+4]}{(\sqrt[3]{162x^3+2})^2+2\sqrt[3]{162x^3+2}+4}$$=\frac{(\sqrt{27x^2-9x+1}-1)(\sqrt{27x^2-9x+1}+1)}{\sqrt{27x^2-9x+1}+1}$
$\Leftrightarrow \frac{6(3x-1)(9x^2+3x+1)}{(\sqrt[3]{162x^3+2}^2+2\sqrt[3]{162x^3+2}+2}=\frac{(3x-1)9x}{\sqrt{27x^2-9x+1}+1}$
$\Leftrightarrow (3x-1)(\frac{18x^2+6x+2}{\sqrt[3]{162x^3+2}^2+2\sqrt[3]{162x^3+2}+4}-\frac{3x}{\sqrt{27x^2-9x+1}+1})=0$
Từ đây suy ra $ x=\frac{1}{3}$ là nghiệm duy nhất của phương trình vì $\frac{18x^2+6x+2}{(\sqrt[3]{162x^3+2})^2+2\sqrt[3]{162x^3+2}+4}-\frac{3x}{\sqrt{27x^2-9x+1}+1}=0$ vô nghiệm
:v ừa :'> –  quangtien1428 26-06-13 01:48 PM
Bạn à cái phương trình cuối vẫn có nghiệm là x=1/3 đấy.... Mình k V được rồi nhé.. ^.^ –  ♂Vitamin_Tờ♫ 26-06-13 09:30 AM

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