a)\(f\left(x\right)=\left(3x+4\right)\cdot\left(5x-1\right)+\left(5x+2\right)\cdot\left(1-3x\right)+2\)

\(=15x^2-3x+20x-4+5x-15x^2+2-6x+2\)

\(=16x\)

b)\(g\left(x\right)=\left(5x-1\right)\cdot\left(2x+3\right)-3\cdot\left(3x-1\right)\)

\(=10x^2+15x-2x-3-9x+3\)

\(=10x^2+4x\)

a, Ta có:

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

\(\Rightarrow\left(3x+4\right)\left(5x-1\right)+\left(5x+2\right)\left(1-3x\right)+2=0\)

\(\Rightarrow15x^2-3x+20x-4+5x-15x^2+2-6x+2=0\)

\(\Rightarrow16x=0-2+4\Rightarrow16x=2\Rightarrow x=\dfrac{1}{8}\)

Vậy nghiệm của đa thức f(x) là \(x=\dfrac{1}{8}\).

b,Ta có:

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

\(\Rightarrow\left(5x-1\right)\left(2x+3\right)-3\left(3x-1\right)=0\)

\(\Rightarrow10x^2+15x-2x-3-9x+3=0\)

\(\Rightarrow10x^2+4x=0\)

\(\Rightarrow2x.\left(5x+2\right)=0\Rightarrow x.\left(5x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\5x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Vậy.................

Chúc bạn học tốt!!!

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 $x-\frac{1}{10}=0$

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

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