toán 9: pt vô tỉ

[lalinhtrang]

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ĐĂNG BÀI NGAY để cùng trao đổi với các thành viên siêu nhiệt tình & dễ thương trên diễn đàn.

1. $\sqrt{x-\dfrac{1}{x}}$-$\sqrt{1-\dfrac{1}{x}}$=$\dfrac{x-1}{x}$
2. $\sqrt{4x^2+5x+1}$+2$\sqrt{x^2+x+1}$=x+3
3. $\sqrt{2x^2+11x+15}$+$\sqrt{x^2+2x-3}$=x+6
4. $\dfrac{1}{x^2}$+$\sqrt{x+5}$=$\dfrac{1}{x}$+$\sqrt{2x+4}$

 

[quynhsieunhan]

$4, \sqrt{4x + 1} - \sqrt{2x + 3} = x - 1$

đk:....................
Đặt $\left\{ \begin{array}{l} \sqrt{4x + 1} = a \\ \sqrt{2x + 3} = b \end{array} \right.$ $(a, b$ \geq $0)$
\Rightarrow Có hệ: $\left\{ \begin{array}{l} 2b^2 - a^2 = 5 (1) \\ a - b = \frac{a^2 - b^2}{2} (2) \end{array} \right.$
$(2)$ \Leftrightarrow $a - b = \frac{(a - b)(a + b)}{2}$ \Rightarrow [TEX]\left[\begin{a - b = 0}\\{a + b = 2} [/TEX]
TH1: $a = b$ \Rightarrow $x = 1$ thay vào (1) thỏa mãn
TH2: $a = 2 - b$ thay vào (1)
\Rightarrow ...........

 

[hien_vuthithanh]

$$\sqrt{x-\dfrac{1}{x}}-\sqrt{1-\dfrac{1}{x}}=\dfrac{x-1}{x}$$

Đk : $\begin{bmatrix}& x\ge 1 & \\ & -1\le x < 0 &\end{bmatrix}$

$\sqrt{x-\dfrac{1}{x}}-\sqrt{1-\dfrac{1}{x}}=\dfrac{x-1}{x}$

$\leftrightarrow \sqrt{\dfrac{x^2-1}{x}}-\sqrt{\dfrac{x-1}{x}}=\dfrac{x-1}{x}$ (*)

$♦ x\ge 1 \rightarrow$ (*) $\leftrightarrow \sqrt{x-1}(\sqrt{\dfrac{x+1}{x}}-\dfrac{1}{\sqrt{x}}-\dfrac{\sqrt{x-1}}{x})=0$

$\leftrightarrow \begin{bmatrix}& x=1 & \\ & \sqrt{\dfrac{x-1}{x}}-\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x-1}}{x} & \end{bmatrix}$

$\sqrt{\dfrac{x-1}{x}}-\dfrac{1}{\sqrt{x}} =\dfrac{\sqrt{x-1}}{x}$

$\leftrightarrow \sqrt{x^2+x}-\sqrt{x}=\sqrt{x-1}$

$\leftrightarrow x^2+x=2x-1+2\sqrt{x^2-x}$

$\leftrightarrow (\sqrt{x^2-x}-1)^2=0$

$\leftrightarrow x^2-x-1=0$

$\leftrightarrow x=\dfrac{1+\sqrt{5}}{2}$

$♦-1\le x<0 \rightarrow$ (*) $\leftrightarrow \sqrt{1-x}(\sqrt{\dfrac{x+1}{-x}}-\dfrac{1}{\sqrt{-x}}-\dfrac{\sqrt{1-x}}{x})=0$

TT TH trên

 

[hien_vuthithanh]

2. $$\sqrt{4x^2+5x+1}+2\sqrt{x^2+x+1}=x+3$$

Đk :$$ \begin{bmatrix}& x\le -1 & \\ & x \ge \dfrac{-1}{4} &\end{bmatrix}$$
$$\sqrt{4x^2+5x+1}+2\sqrt{x^2+x+1}=x+3$$
$$\sqrt{4x^2+5x+1}=a,\sqrt{4x^2+4x+4}=b (a\ge 0 ,b>0)$$
$$\rightarrow \left\{\begin{matrix}& a+b=x+3 & \\ & a^2-b^2=x-3 &
\end{matrix}\right. \leftrightarrow \left\{\begin{matrix}
& a+b=x+3 & \\ & a-b= \dfrac{x-3}{x+3} & \end{matrix}\right. \rightarrow a=\dfrac{x^2+7x+6}{2(x+3)}$$

$$\rightarrow \dfrac{(x^2+7x+6)^2}{4(x+3)^2}=4x^2+5x+1(=a^2)$$
$$\leftrightarrow (x+1)^2(x+6)^2=4(x+3)^2(x+1)(4x+1)$$
$$\leftrightarrow \begin{bmatrix}& x=-1 & \\ & 15x^3+87x^2+102x=0 & \end{bmatrix}$$
$$\leftrightarrow \begin{bmatrix}& x=-1 & \\ & x=0 & \\& x=\dfrac{-29 \pm \sqrt{41}}{10}\end{bmatrix}$$
Thử lại đc $$\rightarrow \begin{bmatrix}& x=-1 & \\ & x=0 & \end{bmatrix}$$

Gop y: xet them truong hop $x=-3$

 

[hien_vuthithanh]

3. $$\sqrt{2x^2+11x+15}+\sqrt{x^2+2x-3}=x+6 $$

Dk :$ \begin{bmatrix}& x\le -3 & \\ & x\ge 1 & \end{bmatrix}$

$\sqrt{2x^2+11x+15}+\sqrt{x^2+2x-3}=x+6$

$\leftrightarrow \sqrt{(x+3)(2x+5)}+\sqrt{(x+3)(x-1)}=x+6 $(*)

$♦ x\ge 1 \rightarrow$ (*) $\leftrightarrow \sqrt{x+3}(\sqrt{2x+5}+\sqrt{x-1})=x+6$

Đặt $ \sqrt{2x+5}=a, \sqrt{x-1}=b (a,b\ge 0)$

$\rightarrow$ (*) $\leftrightarrow \sqrt{a^2-b^2-3}(a+b)=a^2-b^2$

$\leftrightarrow (a+b)(\sqrt{a^2-b^2-3}-(a-b))=0$

$\leftrightarrow \begin{bmatrix}& a+b=0 & \\ & \sqrt{a^2-b^2-3}=(a-b) & \end{bmatrix}$

$\leftrightarrow \begin{bmatrix}& \sqrt{2x+5}= \sqrt{x-1}= 0 (Vô lý)& \\ & \left\{\begin{matrix} & \sqrt{2x+5}\ge \sqrt{x-1} & \\
& a^2-b^2-3=a^2+b^2-2ab &
\end{matrix}\right. &
\end{bmatrix} \leftrightarrow \left\{\begin{matrix}& x \ge -6 & \\ & 2b^2-2ab+3=0 & \end{matrix}\right.$

$\leftrightarrow \left\{\begin{matrix}& x \le -6 & \\ & 2(x-1)-2\sqrt{(2x+5)(x-1)}+3=0 & \end{matrix}\right. \leftrightarrow\begin{bmatrix}& x\ge -6 & \\ & 2\sqrt{(2x+5)(x-1)}=2x+1 & \end{bmatrix} $

Bình phương tiếp.

$♦x\le -3 \rightarrow$ (*)$ \leftrightarrow \sqrt{-x-3}(\sqrt{-2x-5}+\sqrt{1-x})=x+6$

TT TH trên.