Question

Giải các pt sau bằng cách đặt ẩn phụ

a) 3(x²+\(\frac{1}{x^2}\)) - 16(x+\(\frac{1}{x}\))+26=0

b) (x+2)(x+3)(x+8)(x+12)=4

Answer 1

a/ Đặt \(x+\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(\Leftrightarrow3\left(a^2-2\right)-16a+26=0\)

\(\Leftrightarrow3a^2-16a+20=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=2\\x+\frac{1}{x}=\frac{10}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-2x+1=0\\3x^2-10x+3=0\end{matrix}\right.\)

b/ \(\Leftrightarrow\left(x+2\right)\left(x+12\right)\left(x+3\right)\left(x+8\right)=4\)

\(\Leftrightarrow\left(x^2+14x+24\right)\left(x^2+11x+24\right)=4\)

Đề thiếu ko bạn? Vế phải là 4 hay \(4x^2\)?