Câu hỏi của Biển Vũ Đức

Question

Giải phương trình :

8{x+1/x}^2+4{x^2+1/x^2}^2-4{x^2+1/x^2}{x+1/x}^2=(x+4)^2

ngonhuminh · Accepted Answer

$x+\dfrac{1}{x}=a;x^2+\dfrac{1}{x^2}=a^2-2$

$8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)a^2=\left(x+4\right)^2$

$8a^2+4\left(a^2-2\right)\left[\left(a^2-2\right)-a^2\right]=\left(x+4\right)^2$

$8a^2-8\left(a^2-2\right)=\left(x+4\right)^2$

$16=\left(x-4\right)^2;\left[{}\begin{matrix}x-4=4;x=8\x-4=-4;x=0\end{matrix}\right.$Câu trả lời: $x+\dfrac{1}{x}=a;x^2+\dfrac{1}{x^2}=a^2-2$

$8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)a^2=\left(x+4\right)^2$

$8a^2+4\left(a^2-2\right)\left[\left(a^2-2\right)-a^2\right]=\left(x+4\right)^2$

$8a^2-8\left(a^2-2\right)=\left(x+4\right)^2$

$16=\left(x-4\right)^2;\left[{}\begin{matrix}x-4=4;x=8\x-4=-4;x=0\end{matrix}\right.$

Violympic toán 8