Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 10: Giải phương trình: `{x}/{\sqrt{2x+1}-1}=x-2`

Toán Lớp 10: Giải phương trình: {x}/{\sqrt{2x+1}-1}=x-2

Comments ( 2 )

  1. $\dfrac{x}{\sqrt{2x + 1} – 1}$ = x – 2
    ĐKXĐ: x >= -1/2; x $\neq$ 0
    PT <=> $\dfrac{x(\sqrt{2x + 1} + 1)}{(\sqrt{2x + 1} + 1)(\sqrt{2x + 1} – 1)}$ = x – 2
    ⇔ $\dfrac{x(\sqrt{2x + 1} + 1)}{(\sqrt{2x + 1})^2 – 1^2}$ = x – 2
    ⇔ $\dfrac{x(\sqrt{2x + 1} + 1)}{2x + 1 – 1}$ = x – 2
     ⇔ $\dfrac{x(\sqrt{2x + 1} + 1)}{2x}$ = x – 2
    ⇔ $\dfrac{\sqrt{2x + 1} + 1}{2}$ = x – 2
    ⇔ $\sqrt{2x + 1}$ + 1 = 2(x – 2)
    ⇔ $\sqrt{2x + 1}$ = 2x – 4 – 1
    ⇔ $\sqrt{2x + 1}$ = 2x + 1 – 6
    Đặt: $\sqrt{2x + 1}$ = a, ta có:
    PT ⇔ a = a^2 – 6
    ⇔ a^2 – a – 6 = 0
    ⇔ a^2 – 3a + 2a – 6 = 0
    ⇔ a(a – 3) + 2(a – 3) = 0
    ⇔ (a – 3)(a + 2) = 0
    Vì: $\sqrt{2x + 1}$ ≥ 0 ∀ x
    ⇒ $\sqrt{2x + 1}$ + 2 > 0
    ⇒ a – 3 = 0
    ⇔ a = 3
    ⇒ $\sqrt{2x + 1}$ = 3
    ⇔ 2x + 1 = 9
    ⇔ x = 4 (thỏa mãn)
     

  2. Giải đáp:
    \(S = \{4\}\) 
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    \quad \dfrac{x}{\sqrt{2x + 1} – 1} = x – 2\qquad (*)\\
    ĐK:\begin{cases}2x + 1 \geqslant 0\\\sqrt{2x +1}\ne 1\end{cases}\Leftrightarrow \begin{cases}x\geqslant – \dfrac12\\x\ne 1\end{cases}\\
    (*)\Leftrightarrow \dfrac{x\left(\sqrt{2x + 1} + 1\right)}{\left(\sqrt{2x +1} -1\right)\left(\sqrt{2x + 1} + 1\right)} = x – 2\\
    \Leftrightarrow \dfrac{\sqrt{2x+1} + 1}{2} = x -2\\
    \Leftrightarrow \sqrt{2x+1} = 2x – 5\\
    \Leftrightarrow \begin{cases}2x – 5 \geqslant 0\\2x +1 = 4x^2 – 20x + 25\end{cases}\\
    \Leftrightarrow \begin{cases}x \geqslant \dfrac52\\\left[\begin{array}{l}x = \dfrac32\\x = 4\end{array}\right.\end{cases}\\
    \Leftrightarrow x = 4\\
    \text{Vậy}\ S = \{4\}
    \end{array}\) 

Leave a reply

222-9+11+12:2*14+14 = ? ( )