Question

Cho phương trình \(\left(a^2+b^2+c^2+1\right)x-\left(ab+bc+ca\right)=0\), \(\left(a,b,c\in R\right)\)

Nghiệm \(x_0\) của phương trình này thỏa mãn điệu kiện:

\(A.1\le x_0< 2\)

\(B.\left|x_0\right|\ge1\)

\(C.\left|x_0\right|< 1\)

D.\(0< x_0< 1\)

Answer 1

\(\left(a^2+b^2+c^2+1\right)x=ab+bc+ca\)

\(\Leftrightarrow x=\dfrac{ab+bc+ca}{a^2+b^2+c^2+1}\)

Ta có:

\(x^2-1=\dfrac{\left(ab+bc+ca\right)^2}{\left(a^2+b^2+c^2+1\right)^2}-1=\dfrac{\left(ab+bc+ca-a^2-b^2-c^2-1\right)\left(ab+bc+ca+a^2+b^2+c^2+1\right)}{\left(a^2+b^2+c^2+1\right)^2}\)

\(=\dfrac{\left[-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2-2\right]\left[\left(a+b+c\right)^2+a^2+b^2+c^2+2\right]}{4\left(a^2+b^2+c^2+1\right)^2}< 0\)

\(\Rightarrow x^2-1< 0\Rightarrow\left|x\right|< 1\)