Toán Lớp 8: $\frac{1}{x}$ + $\frac{1}{\sqrt[]{2-x^2}}$ = 2
Leave a reply
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.
Toán Lớp 8: $\frac{1}{x}$ + $\frac{1}{\sqrt[]{2-x^2}}$ = 2
Comments ( 2 )
Đây nha
1/x + 1/\sqrt{2 – x^2} = 2
ĐKXĐ: $\begin{cases} 2 – x^2 > 0\\x \neq 0\\\end{cases}$
⇔ $\begin{cases} – \sqrt{2} < x < \sqrt{2}\\x \neq 0\\ \end{cases}$
Đặt: \sqrt{2 – x^2} = a \ (a ≥ 0)
Ta có:
PT ⇔ 1/x + 1/a = 2
⇔ (x + a)/(ax) = 2
⇔ a + x = 2ax \(1)
Có: \sqrt{2 – x^2} = a
⇒ a^2 = 2 – x^2
⇔ a^2 + x^2 = 2 \(2)
⇔ (a + x)^2 – 2ax = 2
Thay \(1) vào \(2), ta có:
(a + x)^2 – (a + x)= 2
⇔ (a + x)^2 – (a + x) – 2 = 0
⇔ (a + x)^2 + (a + x) – 2(a + x) – 2 = 0
⇔ (a + x)(a + x + 1) – 2(a + x + 1) = 0
⇔ (a + x + 1)(a + x – 2) = 0
⇔ $\left[\begin{matrix} a + x = – 1 \\ a + x = 2\end{matrix}\right.$
⇔ $\left[\begin{matrix} a = – 1 – x\\ a = 2 – x\end{matrix}\right.$
⇔ $\left[\begin{matrix} \sqrt{2 – x^2} = – 1 – x\\ \sqrt{2 – x^2} = 2 – x\end{matrix}\right.$
⇔ $\left[\begin{matrix} \begin{cases} – \sqrt{2} < x ≤ – 1\\(\sqrt{2 – x^2}\ )^2 = (- 1 – x)^2\\ \end{cases}\\ \begin{cases} – \sqrt{2} < x < \sqrt{2}\\(\sqrt{2 – x^2}\ )^2 = (2 – x)^2\\ \end{cases}\end{matrix}\right.$
⇔ $\left[\begin{matrix} \begin{cases} – \sqrt{2} < x ≤ – 1\\2 – x^2 = x^2 + 2x + 1\\ \end{cases}\\ \begin{cases} – \sqrt{2} < x < \sqrt{2}\\2 – x^2 = x^2 – 4x + 4\\ \end{cases}\end{matrix}\right.$
⇔ $\left[\begin{matrix} \begin{cases} – \sqrt{2} < x ≤ – 1\\2x^2 + 2x – 1 = 0\\ \end{cases}\\ \begin{cases} – \sqrt{2} < x < \sqrt{2}\\2x^2 – 4x + 2 = 0\\ \end{cases}\end{matrix}\right.$
+) TH_1: 2x^2 + 2x – 1 = 0
⇔ 2(2x^2 + 2x – 1) = 0
⇔ 4x^2 + 4x – 2 = 0
⇔ (2x + 1)^2 – 3 = 0
⇔ (2x + 1)^2 – (\sqrt{3})^2 = 0
⇔ (2x + 1 – \sqrt{3})(2x + 1 + \sqrt{3}) = 0
⇔ $\left[\begin{matrix} x = \dfrac{\sqrt{3} – 1}{2} \text{(Loại)}\\ x = \dfrac{- \sqrt{3} – 1}{2} \text{(TM)}\end{matrix}\right.$
+) TH_2: 2x^2 – 4x + 2 = 0
⇔ 2(x^2 – 2x + 1) = 0
⇔ 2(x – 1)^2 = 0
⇔ x = 1 $\text{(TM)}$
Vậy nghiệm của PT là: S = {\frac{- \sqrt{3} – 1}{2}; 1}