Vì pt có hai nghiệm x1 ; x2 ,Theo hệ thức viet :
\(\int^{x1+x2=-2k}_{x1x2=4}\)
Dễ thấy x1 = x2 = 0 pt vô nghiệm .
TA có : \(\left(\frac{x1}{x2}\right)^2+\left(\frac{x2}{x1}\right)^2\ge3\)
<=> \(\frac{x1^4+x2^4}{x1^2x2^2}\ge3\) <=> \(x1^4+x2^4\ge3x1^2x2^2\left(vìx1;x2khac0\right)\)
<=> \(\left(x1^2+x2^2\right)^2\ge5x1^2x2^2\) <=> \(\left[\left(x1+x2\right)^2-2x1x2\right]^2\ge5x1^2x2^2\)
<=> \(\left[\left(-2k\right)^2-2\cdot4\right]^2\ge5\cdot4^2\) <=> \(l4k^2+8l\ge4\sqrt{5}\)
<=> \(4k^2+8\ge4\sqrt{5}\) hoặc \(4k^2+8\le4\sqrt{5}\)
Số lẻ quá chắc giải sai
bai thi .....................kho..........................kho..............troi.................thilanh.............................ret..................wa.........................dau................wa......................tich....................ung.....................ho.....................cho............do.................lanh
bai thi .....................kho..........................kho..............troi.................thilanh.............................ret..................wa.........................dau................wa......................tich....................ung.....................ho.....................cho............do.................lanh