1, $(\sqrt{x-1}+1)^{3}+2\sqrt{x-1}=2-x$
2, $\sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{5x}$
3, $6x^{2}-3\sqrt{3x^{2}-2x-1}=4(x+1)$
4, $\sqrt{x^{2}-1}+\sqrt{x^{2}-3x+2}=\sqrt{x^{2}-x}$
5, $x+\sqrt{4-x^{2}}=2+3x\sqrt{4-x^{2}}$
a) $$\sqrt{x-1}(3\sqrt{x-1}+x+5)=0$$
b) Nhân liên hợp:
$$\left(\sqrt[3]{x+1}-\dfrac{\sqrt[3]{4}}{4}(2x+1)\right)+\left(\sqrt[3]{x+1}-\dfrac{\sqrt[3]{4}}{4}(2x-1)\right)-\left(\sqrt[3]{5x}-\frac{\sqrt[3]{4}+1}{2} x\right)=0$$
c) $$ \left( \sqrt {3\,{x}^{2}-2\,x-1}-2 \right) \left( 2\,\sqrt {3\,{x}^{2}-2\,x-1}+1 \right) =0$$
d) $$\sqrt{x+1}\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}+\sqrt{x-2}\right)=0$$
e) $$\left( 3\,\sqrt {4-{x}^{2}}+3\,x+4 \right) \left( -2+x+\sqrt {4-{x}^{2}} \right) =0$$
Bài viết đã được chỉnh sửa nội dung bởi nthoangcute: 26-03-2013 - 22:15