1)) $\frac{1001x^{4}+x^{4}\sqrt{2x^{2}+2002}+4x^{2}}{999}=2002$
2)) $x^{2}-2(x+1)\sqrt{x^{2}-1}-3x^{2}+6x-1=0$
3)) $4x^{4}+x^{2}+3x+4=3.\sqrt[3]{16x^{3}+2x}$
4)) $\sqrt{x}+\sqrt{1-x}+\sqrt{1+x}=2$
1) $\Leftrightarrow 1001x^4+x^4\sqrt{2x^2+2002}+4x^2=2002.999$
$\Leftrightarrow x^4(1001+\sqrt{2x^2+2002})+2[(2x^2+2002)-1001^2]=0$
$\Leftrightarrow (\sqrt{2x^2+2002}+1001)[x^4+2(\sqrt{2x^2+2002}-1001)]=0$
$\Rightarrow x^4+2(\sqrt{2x^2+2002}-1001)=0$
$\Leftrightarrow x^4+2\sqrt{2x^2+2002}=2002$
$\Leftrightarrow (x^4+2x^2+1)-(2x^2+2002-2\sqrt{2x^2+2002}+1)=0$
$\Leftrightarrow (x^2+1)^2-(\sqrt{2x^2-1})^2=0$
............. DỄ RỒI.........
2) ĐKXĐ: $x\leq -1, x \geq 1$
PT $\Leftrightarrow (x^2-1)-2(x+1)\sqrt{x^2-1}+(x^2+2x+1)=4x^2-4x+1$
$\Leftrightarrow (\sqrt{x^2-1}-x-1)^2=(2x-1)^2$
............Đến đây dễ rồi...
Bài viết đã được chỉnh sửa nội dung bởi Nguyen Minh Hai: 25-02-2015 - 01:13