a ) \(4x^2+2x+1=\left(2x\right)^2+2\cdot2x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

b ) \(x^2+3x+4=\left(x^2+2\cdot\frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{7}{4}=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)

c ) \(9x^2+3x+5=\left(3x\right)^2+2\cdot3x\cdot\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(3x+\frac{1}{2}\right)^2+\frac{19}{4}>0\forall x\)

Ta có : 4x² + 2x + 1

= (2x)² + 2.2x.\(\frac{1}{2}\) + \(\frac{1}{2}+\frac{3}{4}\)

= (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)

Mà : (2x + \(\frac{1}{2}\))² \(\ge0\forall x\)

=> (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) \(\ge\frac{3}{4}\forall x\)

Hay : (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)  \(>0\forall x\)

Vậy 4x² + 2x + 1 \(>0\forall x\)

1.

x(x+1)(x²+x+3) = (x²+x)(x²+x+3)

đặt x²+x = t

=> t(t+3)=4

=>t;t+3 thuộc Ư(4)

=> t;t+3 thuộc -1;1-2;2-4;4

tự xét lần lượt các TH nha bạn

Ta có:\(-x^2+4x-7\)

\(=-\left(x^2-4x+7\right)\)

\(=-\left(x^2-2.x.2+2^2-4+7\right)\)

\(=-\left[\left(x-2\right)^2+3\right]\)

\(=-\left(x-2\right)^2-3\)

Do \(-\left(x-2\right)^2\le0\) với \(\forall x\)

\(\Rightarrow-\left(x-2\right)^2-3\le-3< 0\)

\(\Rightarrow-x^2+4x-7< 0\) (đpcm)

câu b,c đề sai bạn nhé!

a , Ta có \(x^2+x+1=x^2+2x\frac{1}{2}+\left(\frac{1}{2}\right)^2+\)\(\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\) \(\ge\frac{3}{4}>0\left(đpcm\right)\)

b , Ta có : \(4x^2-2x+3\)= \(\left(2x\right)^2-2.2x.1+1^2+2\) = \(\left(2x-1\right)^2+2\ge2>0\left(đpcm\right)\)

c , Ta có \(3x^2+2x+1=x^2-\frac{2x}{3}+\frac{1}{9}+2x^2+\frac{8x}{3}+\frac{8}{9}\)

= \(\left(x-\frac{1}{3}\right)^2+2\left(x^2+\frac{4x}{3}+\frac{4}{9}\right)=\left(x-\frac{1}{3}\right)^2+2\left(x+\frac{2}{3}\right)^2\ge0\)

Vì Dấu "=" không thể xảy ra , do đó \(3x^2+2x+1>0\left(đpcm\right)\)

a,-x²+x+1>0 với mọi x mới đúng

anh gioi qua

hơi ngán dạng này :((((

a, \(x^2-3x+5=x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+5=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

b,

\(x^2-\frac{1}{3}x+\frac{5}{4}=x^2-2.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{5}{4}=\left(x-\frac{1}{6}\right)^2+\frac{11}{9}>0\forall x\)

c,

\(x-x^2-3=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}-3=-\left(x-\frac{1}{2}\right)^2-\frac{11}{4}< 0\forall x\)d,

\(x-2x^2-\frac{5}{2}=-2\left(x^2-\frac{1}{2}x+\frac{5}{4}\right)=-2\left(x^2-2.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{5}{4}\right)=-2\left[\left(x-\frac{1}{4}\right)^2+\frac{19}{16}\right]=-2\left(x-\frac{1}{4}\right)^2-\frac{19}{8}< 0\forall x\)P/s : ko chắc lém :)))

\(f\left(x\right)=4x^3+4x^4-x^2+3x^2-3x^4-3x^3\)

\(\Leftrightarrow f\left(x\right)=\left(4x^3-3x^3\right)+\left(4x^4-3x^4\right)+\left(-x^2+3x^2\right)\)

\(\Leftrightarrow f\left(x\right)=x^3+x^4+2x^2\)

\(f\left(x\right)=0\)

\(\Leftrightarrow x^3+x^4+2x^2=0\)

\(\Leftrightarrow x^2\left(x+x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)+\dfrac{3}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\forall x\end{matrix}\right.\)

Vậy f(x) chỉ có 1 nghiệm