\(P=16x^2+8x+2=\left(16x^2+8x+1\right)+1=\left(4x+1\right)^2+1\)

Do \(\left\{{}\begin{matrix}\left(4x+1\right)^2\ge0\\1>0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow P=\left(4x+1\right)^2+1>0;\forall x\) (đpcm)

\(P=16x^2+8x+2\)

\(=\left(16x^2+8x+1\right)+1\)

\(=\left[\left(4x\right)^2+2\cdot4x\cdot1+1^2\right]+1\)

\(=\left(4x+1\right)^2+1\)

Ta thấy: \(\left(4x+1\right)^2\ge0\forall x\)

\(\Leftrightarrow P=\left(4x+1\right)^2+1\ge1>0\forall x\)

hay \(P\) luôn dương với mọi \(x\).

C=(4x)²+4x+1+99

=(4x+1)²+99>0

Vậy biểu thức luôn dương

Chúc hok tốt

Xét \(C=16x^2+4x+100\)

\(C=4x\left(4x+1\right)+100\)

Mà \(4x\left(4x+1\right)\ge0,\forall x\)( \(\forall x\)nghĩa là VỚI MỌI X nha bạn)

\(\Rightarrow4x\left(4x+1\right)+100>0,\forall x\)

\(\Leftrightarrow C>0\)

 Vậy, \(4x\left(4x+1\right)+100>0,\forall x\)(ĐPCM)

Cách 2: 

Xét \(C=16x^2+4x+100\)

\(C=\left(4x\right)^2+2x\cdot2+4+96\)

\(C=\left(4x\right)^2+2x\cdot2+2^2+96\)

\(C=\left(4x+2\right)^2+96\)

Vì \(\left(4x+2\right)^2\ge0,\forall x\)

\(\Rightarrow\left(4x+2\right)^2+96>0,\forall x\)

\(\Leftrightarrow C>0\)

Vậy...

đưa nó về dạng 

nếu luôn âm : -ax²+b

ne6i1 luôn dương : ax²+b

A=(x+2)^2 +3

B=(x-5)^2 +4

C=4(x+1/2)^2 +4

D=(x-1/2)^2 +19/4

E=2(x-3/4)^2 +95/8

\(f,F=x^2+9y^2-8x+4y+27\) (sửa đề)

\(=\left(x^2-8x+16\right)+\left(9y^2+4y+\dfrac{4}{9}\right)+\dfrac{95}{9}\)

\(=\left(x^2-2\cdot x\cdot4+4^2\right)+\left[\left(3y\right)^2+2\cdot3y\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2\right]+\dfrac{95}{9}\)

\(=\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2+\dfrac{95}{9}\)

Ta thấy: \(\left(x-4\right)^2\ge0\forall x\)

             \(\left(3y+\dfrac{2}{3}\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2+\dfrac{95}{9}\ge\dfrac{95}{9}>0\forall x;y\)

hay \(F\) luôn dương với mọi \(x;y\).

\(Toru\)

Ta có :

\(B=x^2-10x+28\)

\(\Rightarrow B=x^2-2.x.5+25+3\)

\(\Rightarrow B=\left(x+5\right)^2+3\)

Vì \(\left(x+5\right)\ge0\) ( với mọi x )

\(\Rightarrow\left(x+5\right)+3\ge3\)

=> đpcm

\(A=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)

\(B=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)

\(C=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)

\(D=\left(x-5\right)^2+\left(3y+1\right)^2+4\)

\(E=\left(4x+1\right)^2+\left(y-2\right)^2+1\)

\(M=-\left(x+\frac{7}{2}\right)^2-\frac{11}{4}\)

\(N=-5\left(x-\frac{3}{5}\right)^2-\frac{41}{5}\)

\(C\) đề sai ví dụ \(x=3\Rightarrow C=2>0\)

\(D=-5\left(x-\frac{7}{10}\right)^2-\frac{131}{20}\)

a)\(A=x^2+x+1=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

b) \(B=2x^2+2x+1=2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{2}=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\)