Chủ đề này có 1 trả lời

[phamquyen134]

1/ Giải pt:

a/ $5x^{2}-(3x+1)\sqrt{2x^{2}+3}-\frac{1}{2}x+3=0$

b/ $(x-2)(\sqrt{3x+1}-1)=3x$

c/ $(x+1)\sqrt{2x^{2}-2x}=2x^{2}-3x-2$

d/ $3\sqrt{x^{2}+4x-5}+\sqrt{x-3}=\sqrt{11x^{2}+25x+2}$

[githenhi512]

1/ Giải pt:

a/ $5x^{2}-(3x+1)\sqrt{2x^{2}+3}-\frac{1}{2}x+3=0$

b/ $(x-2)(\sqrt{3x+1}-1)=3x$

c/ $(x+1)\sqrt{2x^{2}-2x}=2x^{2}-3x-2$

d/ $3\sqrt{x^{2}+4x-5}+\sqrt{x-3}=\sqrt{11x^{2}+25x+2}$

d. ĐK: $x\geq 3$

$\Leftrightarrow 9(x^2+4x-5)+x-3+6\sqrt{(x^2+4x-5)(x-3)}=11x^2+25x+2\Leftrightarrow 6\sqrt{(x^2+4x-5)(x-3)}=2x^2-12x+50$

Đặt $\sqrt{x^2+4x-5}=a> 0, \sqrt{x-3}=b\geq 0\Rightarrow 6ab=2a^2-20b^2\Leftrightarrow a^2-10b^2-3ab=0\rightarrow a=5b\Rightarrow \sqrt{x^2+4x-5}=5\sqrt{x-3}\Rightarrow ..$

c. Đk: $x\geq 1 hoặc x\leq 0$

$Đặt \sqrt{2x^2-2x}=a\geq 0\Rightarrow (x+1)a=a^2-x-2\Leftrightarrow (a+1)(a-2-x)=0\Rightarrow a=x+2\Leftrightarrow \sqrt{2x^2-2x}=x+2\Rightarrow ...$

b. Đk: $x\geq \frac{-1}{3}$

$\Leftrightarrow (x-2).\frac{3x}{\sqrt{3x+1}+1}=3x$

N x=0(t/m)

N $x\neq 0\Rightarrow x-2=\sqrt{3x+1}+1\Leftrightarrow \sqrt{3x+1}=x-3\Rightarrow ...$