Giải:

a) \(F\left(x\right)+G\left(x\right)-H\left(x\right)\)

\(=4x^2+3x-2+3x^2-2x+5-\left[x\left(5x-2\right)+3\right]\)

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

\(=4x^2+3x-2+3x^2-2x+5-5x^2+2x-3\)

\(=2x^2+3x\)

Để \(F\left(x\right)+G\left(x\right)-H\left(x\right)=0\)

\(\Leftrightarrow2x^2+3x=0\)

\(\Leftrightarrow x\left(2x+3\right)=0\)

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

b) \(F\left(x\right)-3x+5\)

\(=4x^2+3x-2-3x+5\)

\(=4x^2+3\)

Vì \(x^2\ge0;\forall x\)

\(\Leftrightarrow4x^2\ge0;\forall x\)

\(\Leftrightarrow4x^2+3\ge3>0;\forall x\)

Vậy ...

a)\(f\left(x\right)=x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-\frac{1}{4}x+2x-3\)

\(=x^5-x^5+7x^4-9x^3-3x^2+2x^2+x^2-\frac{1}{4}x+2x-3\)

\(=7x^4-9x^3+\frac{7}{4}x-3\)

\(g\left(x\right)=5x^4-x^5+\frac{1}{2}x^2+x^5+x^2-4x^4-2x^3+3x^2+x^3-\frac{1}{4}\)

\(=-x^5+x^5+5x^4-4x^4-2x^3+x^3+\frac{1}{2}x^2+x^2+3x^2-\frac{1}{4}\)

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

b)\(f\left(1\right)=7.1^4-9.1^3+\frac{7}{4}.1-3=7-9+\frac{7}{4}-3=-\frac{13}{4}\)

\(f\left(-1\right)=7.\left(-1\right)^4-9.\left(-1\right)^3+\frac{7}{4}.\left(-1\right)-3=7+9-\frac{7}{4}-3=\frac{45}{4}\)

\(g\left(1\right)=1^4-1^3+\frac{9}{2}.1^2-\frac{1}{4}=1-1+\frac{9}{2}-\frac{1}{4}=\frac{17}{4}\)

\(g\left(-1\right)=\left(-1\right)^4-\left(-1\right)^3+\frac{9}{2}.\left(-1\right)^2-\frac{1}{4}=1+1+\frac{9}{2}-\frac{1}{4}=\frac{25}{4}\)

c) Ta có: f(x)+g(x)=\(7x^4-9x^3+\frac{7}{4}x-3+x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}=7x^4+x^4-9x^3-x^3+\frac{9}{2}x^2+\frac{7}{4}x-3-\frac{1}{4}\)

\(=8x^4-10x^3+\frac{9}{2}x^2+\frac{7}{4}x-\frac{13}{4}\)

f(x)-g(x) =\(7x^4-9x^3+\frac{7}{4}x-3-x^4+x^3-\frac{9}{2}x^2+\frac{1}{4}=7x^4-x^4-9x^3+x^3-\frac{9}{2}x^2+\frac{7}{4}x-3+\frac{1}{4}\)

\(=6x^4-8x^3-\frac{9}{2}x^2+\frac{7}{4}x-\frac{11}{4}\)

mi bn oi help me nha

Làm tắt thôi nhé bn !

Có h(x) = f (x) + g (x) = 3x² + 2 ( sau khi tính kết quả sẽ ra vậy nhé ! mk làm tắt )

Lại có h ( x) có :

3x²  \(\ge\)0

2 >0 

Từ 2 điều này => 3x² +2 \(\ge2\)

=> h(x) ko có nghiệm

          F(x) = \(-6x^3+8x^2-\frac{1}{2}-4^4\)

 +       G(x) =    \(6x^3-5x^2+\frac{5}{2}+4x^4\)

_________________________________________

          H(x) =                  \(3x^2+3\)

Vậy H(x) = 3x² + 3

\(f\left(x\right)+g\left(x\right)=\left(-6x^3+8x^2-\frac{1}{2}-4^4\right)+\left(4x^4+\frac{5}{2}-5x^2+6x^3\right)\)

\(=-6x^3+8x^2-\frac{1}{4}-4^4+4x^4+\frac{5}{2}-5x^2+6x^3\)

\(=\left(-6x^3+6x^3\right)+\left(8x^2-5x^2\right)+\left(-\frac{1}{2}+\frac{5}{2}\right)+\left(-4^4+4x^4\right)\)

\(=3x^2+2\)

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

Ta có: \(3x^2\ge0\forall x\)

\(\Rightarrow3x^2+2\ge0\forall x\)

Vậy: h(x) = 3x² + 2 không có nghiệm

\(f\left(x\right)=\left(x-1\right).g\left(x\right)\)

\(\Rightarrow3x^3-2x^2+x+5=\left(x-1\right)\left(3x^2+ax+b\right)\)

\(\Rightarrow3x^3-2x^2+x+5=3x^3+ax^2+bx-3x^2-ax-b\)

\(\Rightarrow-2x^2+x+5=x^2\left(a-3\right)+x\left(b-a\right)-b\)

-Bạn kiểm tra lại đề.

`K(x)=F(x)+G(x)`

`K(x)=(3x^2+2x-5)+(-3x^2-2x+2)`

`= 3x^2+2x-5-3x^2-2x+2`

`= (3x^2-3x^2)+(2x-2x)+(-5+2)`

`= -3`

Bậc của đa thức: `0`

`@` `\text {dnammv}`

Tìm min:

$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$

Tìm min

$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$

$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)

Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$

$\Leftrightarrow x=\frac{-1}{4}$

Tìm min

$H=5x^2-x+1=5(x^2-\frac{x}{5})+1$

$=5[x^2-\frac{x}{5}+(\frac{1}{10})^2]+\frac{19}{20}$

$=5(x-\frac{1}{10})^2+\frac{19}{20}\geq \frac{19}{20}$
Vậy $H_{\min}=\frac{19}{20}$. Giá trị này đạt tại $x-\frac{1}{10}=0$

$\Leftrightarrow x=\frac{1}{10}$