Question

Cho \(a=\dfrac{x}{x^2-x+1}\). Tính theo a giá trị của biểu thức: \(P=\dfrac{x^2}{x^4+x^2+1}\)

Answer 1

\(P=\dfrac{x^2}{x^4+x^2+1}=\dfrac{x^2}{x^4+2x^2+1-x^2}=\dfrac{x^2}{\left(x^2+1\right)^2-x^2}=\dfrac{x^2}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

\(=a\cdot\dfrac{x}{x^2+x+1}\)

Có \(a=\dfrac{x}{x^2-x+1}\Rightarrow\dfrac{1}{a}=\dfrac{x^2-x+1}{x}=x+\dfrac{1}{x}-1\)

Đặt \(B=\dfrac{x}{x^2+x+1}\Rightarrow\dfrac{1}{B}=\dfrac{x^2+x+1}{x}=x+\dfrac{1}{x}+1=\dfrac{1}{a}-2\)

\(\Leftrightarrow\dfrac{1}{B}=\dfrac{1-2a}{a}\Leftrightarrow B=\dfrac{a}{1-2a}\)

Do đó \(P=a\cdot\dfrac{a}{1-2a}=\dfrac{a^2}{1-2a}\)

 

Answer 2

Hic sao hay lỗi công thức thế :<

Do đó \(\dfrac{1}{B}=\dfrac{1-2a}{a}\Leftrightarrow B=\dfrac{a}{1-2a}\)

\(P=a\cdot\dfrac{a}{1-2a}=\dfrac{a^2}{1-2a}\)

Answer 3

Trần Thanh Phương

1/B = 1/a+2 chứ