$\sqrt[3]{7x+1} -\sqrt[3]{x^{2}-x-8}+ \sqrt[3]{x^{2}-8x-1}=2$ 
$10x^{2} +3x +1=(1+6x)\sqrt{x^{2}+3}$
$\sqrt{\frac{x^{3}}{3-4x}} -\frac{1}{2\sqrt{x} }=\sqrt{x} $
$(3x-5).\sqrt{2x^{2}-3}=4x^{2}-6z+1$
$27\sqrt{5+2x}+27\sqrt{4-2x}\geq(4x+1)^{2}$
bạn ơi xem lại đề câu số 4 –  chaicolovenobita 12-06-13 07:51 AM
câu 2: đặt $a=\sqrt{x^{2}+3}  ta  có  9x^{2}+a^{2}+3x-2=(6x+1)a\Leftrightarrow 9x^{2}-6ax+a^{2}+3x-a-2=0\Leftrightarrow (3x-a)^{2}+(3x-a)-2=0\Leftrightarrow (3x-a-1)(3x-a+2)=0$
câu 1:
$theo  đề  ta  viết  lại:  \sqrt[3]{7x+1}+\sqrt[3]{-x^{2}+x+8}+\sqrt[3]{x^{2}-8x-1}=2.  CM: (a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a).  đặt  a=\sqrt[3]{7x+1},  b=\sqrt[3]{-x^{2}+x+8}, c=\sqrt[3]{x^{2}-8x-1}\Rightarrow \begin{cases}a+b+c=2 \\ a^{3}+b^{3}+c^{3}=8 \end{cases}\Rightarrow (a+b+c)^{3}=a^{3}+b^{3}+c^{3}\Leftrightarrow 3(a+b)(b+c)(c+a)=0,  bạn  tự  giải  tiếp.$
1)$\sqrt[3]{7x+1}-\sqrt[3]{x^{2}-x-8}=2-\sqrt[3]{x^{2}-8x-1}$
$\frac{7x+1-x^{2}+x+8}{\sqrt[3]{(x^{2}-x-8)^{2}}+\sqrt[3]{7x+1^{2}}+\sqrt[3]{(7x+1)(x^{2}-x-1})}=\frac{8-x^{2}+8x+1}{4+\sqrt[3]{(x^{2}-8x-1)^{2}}+2\sqrt[3]{(x^{2}-8x-1)^{2}}}$
$\Rightarrow x^{2}-8x-9=0$
$\Rightarrow (x-9)(x+1)=0$
x=9 hoặc x=-1


đã bảo chỗ còn lại tự làm mà ,nhớ đọc kĩ huongs dẫn sử dụng tr khi dùng –  chaicolovenobita 13-06-13 06:15 AM
bạn làm thiếu nghiệm oy –  thanhgaubong 12-06-13 09:55 PM
bài này chưa làm hoàn chỉnh, phần mẫu bạn tự lập luận, vì mình k yêu cầu trả vỏ sò nên chỉ thế thôi, tùy bạn –  chaicolovenobita 12-06-13 07:39 AM

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