Chủ đề này có 3 trả lời

[tientran1802]

[BysLyl]

Câu 2:

1. Pt (1) có nghiệm khi$\Delta '\geq 0\Rightarrow \Delta '=(m-1)^{2}-(m^{2}-3)=m^{2}-2m+1-m^{2}+3=4-2m\geq 0\Rightarrow m\leq 2$

2. Theo hệ thức Viet $x_{1}=3x_{2}\Rightarrow \left\{\begin{matrix} x_{1}+x_{2}=2m-2\\ x_{1}.x_{2}=m^{2}-3 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} 4x_{2}=2m-2\\ 3x_{2}^{2}=m^{2}-3 \end{matrix}\right. \Rightarrow 3.(\frac{m-1}{2})^{2}=m^{2}-3\Rightarrow 3m^{2}-6m+3=4m^{2}-12\Leftrightarrow m^{2}+6m-15=0\Rightarrow \begin{bmatrix} m=2\sqrt{6}-3> -2\\ m=-2\sqrt{6}-3< -2 \end{bmatrix}$

$m=-2\sqrt{6}-3$

P/s: vô tỉ à. ai kiểm tra giúp mình với

[yeutoan2604]

1a) $P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}=\frac{(x+2)(\sqrt{x}+1)+(\sqrt{x}+1)^{2}(\sqrt{x}-1)-(\sqrt{x}+1)(x+\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{x\sqrt{x}-\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}$

b) Ta có $x\geq 0;x\neq 1\Rightarrow (\sqrt{x}-1)^{2}>0\Leftrightarrow x-2\sqrt{x}+1>0\Leftrightarrow x+\sqrt{x}+1>3\sqrt{x}\Rightarrow \frac{\sqrt{x}}{x+\sqrt{x}+1}<\frac{1}{3}$

[yeutoan2604]

2a) Như trên

b) Theo Vi-ét ta có $\left\{\begin{matrix} x_{1}+x_{2} &=2m-2 \\ x_{1}x_{2} & =m^{2}-3 \end{matrix}\right.\Rightarrow x_{1}=3x_{2}\Leftrightarrow x_{1}+x_{2}=4x_{2}\Leftrightarrow x_{2}=\frac{m-2}{2}\Rightarrow x_{1}=\frac{3m-2}{2}\Rightarrow x_{1}x_{2}=m^{2}-3\Leftrightarrow (m-2)(3m-2)=4(m^{2}-3)\Leftrightarrow -m^{2}-8m+16=0\Leftrightarrow \begin{bmatrix} m &=-4+4\sqrt{2} (t/m)\\ m & =-4-4\sqrt{2}(t/m) \end{bmatrix}$