Question

[Nguyễn Anh Khoa]

E=3x²+x+5 chứng minh biểu thức luôn âm,dương

Answer 1

[Xyz OLM]

Ta có E = \(3x^2+x+5=3\left(x^2+\frac{x}{3}+\frac{5}{3}\right)=3\left(x^2+2.x.\frac{1}{6}+\frac{1}{36}+\frac{59}{36}\right)\)

\(=3\left(x+\frac{1}{6}\right)^2+\frac{59}{12}\ge\frac{59}{12}>0\)

=> E luôn dương với mọi x 

Answer 2

[Quỳnh Anh]

Trả lời:

\(E=3x^2+x+5=3\left(x^2+\frac{1}{3}x+\frac{5}{3}\right)=3\left(x^2+2.x.\frac{1}{6}+\frac{1}{36}+\frac{59}{36}\right)\)

\(=3\left[\left(x+\frac{1}{6}\right)^2+\frac{59}{36}\right]=3\left(x+\frac{1}{6}\right)^2+\frac{59}{12}\)

Ta có: \(\left(x+\frac{1}{6}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(x+\frac{1}{6}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(x+\frac{1}{6}\right)^2+\frac{59}{12}\ge\frac{59}{12}>0\forall x\)

Dấu "=" xảy ra khi x + 1/6 = 0 <=> x = - 1/6

Vậy biểu thức E luôn dương.