Question

1. \(x^4-x+\frac{1}{2}>0\)

2. \(4a^4-4a^3+5a^2-2a+1>0\)

Answer 1

1. \(x^3-x+\frac{1}{2}=x^4-x^2+\frac{1}{4}+x^2-x+\frac{1}{4}=\left(x^2-\frac{1}{2}\right)^2+\left(x-\frac{1}{2}\right)^2\ge0\)

Nếu  \(\left(x^2-\frac{1}{2}\right)^2+\left(x-\frac{1}{2}\right)^2=0\)thì \(\hept{\begin{cases}x-\frac{1}{2}=0\\x^2-\frac{1}{2}=0\end{cases}=>\hept{\begin{cases}x=\frac{1}{2}\\x^2=\frac{1}{2}\end{cases}}}\)(VÔ LÍ)

Vậy \(x^4-x+\frac{1}{2}>0\)

Answer 2

2/ \(BT=a^2\left(4a^2-4a+5\right)-2a+1\)

\(=\left(2a-1\right)^2.a^2+\left(4a^2-2a+1\right)\)

\(=\left(2a^2-a\right)^2+\left(2a-\frac{1}{2}\right)^2+\frac{3}{4}>0\)