Question

Cho \(\frac{x}{x^2+x+1}\)= -\(\frac{2}{3}\) . Tính giá trị biểu thức : P = \(\frac{x^2}{x^4+x^2+1}\)

Answer 1

Sửa một chút nhầm lẫn:

...

\(\Rightarrow x^2-x+1=-2x-\frac{3}{2}x=-\frac{7}{2}x\)

Từ (*) (**)

\(P=\left(-\frac{2}{3}\right)\left(\frac{x}{-\frac{7}{2}x}\right)=\left(-\frac{2}{3}\right)\left(-\frac{2}{7}\right)=\frac{4}{21}\)

Đáp số : P=4/21

Answer 2

\(\frac{x}{x^2+x+1}=-\frac{2}{3}\) (1)

\(P=\frac{x^2}{x^4+x^2+1}\\ \)

Ta có: \(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)

\(\Rightarrow P=\frac{x}{\left(x^2+x+1\right)}.\frac{x}{x^2-x+1}=-\frac{2}{3}.\frac{x}{x^2-x+1}\) (*)

\(\left(1\right)\Rightarrow x^2+x+1=-\frac{3}{2}x\)

\(\Rightarrow x^2-x+1=-2x-\frac{3}{2}x=-\frac{7}{3}x\) (**)

Từ (*)&(**) \(P=\left(-\frac{2}{3}\right)\left(\frac{x}{-\frac{7}{3}x}\right)=\left(-\frac{2}{3}\right)\left(-\frac{3}{7}\right)=\frac{2}{7}\)