a) Ta có: \(x^4-x^2-2=0\)
\(\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra \(x^2-2=0\)
\(\Leftrightarrow x^2=2\)
hay \(x=\pm\sqrt{2}\)
Vậy: \(x=\pm\sqrt{2}\)
b) Ta có: \(\left(x+1\right)^4-\left(x^2+2\right)^2=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2-\left(x^2+2\right)\right]\cdot\left[\left(x+1\right)^2+\left(x^2+2\right)\right]=0\)
\(\Leftrightarrow\left(x^2+2x+1-x^2-2\right)\cdot\left(x^2+2x+1+x^2+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\cdot\left(2x^2+2x+3\right)=0\)(3)
Ta có: \(2x^2+2x+3\)
\(=2\left(x^2+x+\frac{3}{2}\right)\)
\(=2\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{5}{4}\right)\)
\(=2\cdot\left[\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\right]\)
\(=2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\forall x\)
hay \(2x^2+2x+3\ge\frac{5}{2}\ne0\forall x\)(4)
Từ (3) và (4) suy ra \(2x-1=0\)
\(\Leftrightarrow2x=1\)
hay \(x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
a) x4 - x2 - 2 = 0
⇔ x4 - 2x2 + x2 - 2 = 0
⇔ x2(x2 - 2) + (x2 - 2) = 0
⇔ (x2 - 2)(x2 + 1) = 0
Do x2 + 1 > 0 ∀x
=> x2 - 2 = 0
⇔ x2 = 2
⇔ x = \(\pm\sqrt{2}\)
Vậy S = {\(\pm\sqrt{2}\)}
b) (x+1)4 - (x2+ 2)2 = 0
⇔ [(x+1)2 - x2 - 2][(x+1)2 + x2 + 2] = 0
⇔(x2 + 2x + 1 - x2 - 2)(x2 + 2x + 1 + x2 + 2) = 0
⇔ (2x - 1)2(x2 + x + 1) = 0
Do x2 + x + 1 > 0 ∀x
=> 2x - 1 = 0
⇔ 2x = 1
⇔ x = \(\frac{1}{2}\)
Vậy S = {\(\frac{1}{2}\)}
a, x4 - x2 - 2 = 0
\(\Leftrightarrow\) (x2 + \(\frac{1}{2}\))2 - \(\frac{9}{4}\) = 0
\(\Leftrightarrow\) x2 + \(\frac{1}{2}\) = \(\frac{3}{2}\) hoặc x2 + \(\frac{1}{2}\) = -\(\frac{3}{2}\)
Xét TH:
TH1: x2 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
\(\Leftrightarrow\) x2 = 1
\(\Leftrightarrow\) x = 1 (TM)
TH2: x2 + \(\frac{1}{2}\) = -\(\frac{3}{2}\)
\(\Leftrightarrow\) x2 = -2
\(\Leftrightarrow\) x = \(\sqrt{-2}\) (KTM vì căn bậc hai luôn luôn có dạng \(\sqrt{x}\) với x \(\ge\) 0)
Vậy S = {1}
b, (x + 1)4 - (x2 + 2)2 = 0
\(\Leftrightarrow\) [(x + 1)2 + x2 + 2][(x + 1)2 - x2 - 2] = 0
\(\Leftrightarrow\) (x2 + 2x + 1 + x2 + 2)(x2 + 2x + 1 - x2 - 2) = 0
\(\Leftrightarrow\) (2x2 + 2x + 3)(2x - 1) = 0
\(\Leftrightarrow\) 2x2 + 2x + 3 = 0 hoặc 2x - 1 = 0
\(\Leftrightarrow\) x = \(-\frac{1-x\sqrt{5}}{2}\); x = \(-\frac{1+x\sqrt{5}}{2}\); x = \(\frac{1}{2}\) (phần 2x2 + 2x + 3 rất khó phân tích nên tui lấy luôn trên mạng :))
Vậy S = {\(-\frac{1-x\sqrt{5}}{2}\); \(-\frac{1+x\sqrt{5}}{2}\); \(\frac{1}{2}\)}
Chúc bạn học tốt!
