Bạn tự phân tích đa thức thành nhân tử nhé! 

\(1.\)

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

\(\Leftrightarrow\)  \(\left(x+1\right)\left(2x^2-2x+3\right)=0\)  \(\left(1\right)\)

Vì  \(2x^2-2x+3=2\left(x^2-x+1\right)+1=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}>0\)  với mọi  \(x\in R\)

nên từ  \(\left(1\right)\)  \(\Rightarrow\)  \(x+1=0\)  \(\Leftrightarrow\)  \(x=-1\)

1)2x^3+x+3=0=>

Phân tích ra thôi bạn!

\(\dfrac{x}{2}\left(4x-3\right)+2\left(3-x\right)\left(x+4\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2}{2}-\dfrac{3x}{2}+2\left(3x+12-x^2-4x\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x}{2}+6x+24-2x^2-8x\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x+2\left(6x+24-2x^2-8x\right)}{2}\le0\)

\(\Leftrightarrow4x^2-3x+12x+48-4x^2-16x\le0\)

\(\Leftrightarrow-7x\le-48\)

\(\Leftrightarrow x\ge\dfrac{48}{7}\)

=>-7x+48≤0

<=>-7x≤-48

<=>(-7x)(-1)≥(-48)(-1)

<=>\(\dfrac{7x}{7}\)≥\(\dfrac{48}{7}\)

<=>x≥\(\dfrac{48}{7}\)

<=>x(x^2-1)=0

<=>x=0 hoặc x^2-1=0

<=>x=0 hoặc x^2=1

<=>x=0 hoặc x=1 hoặc x=-1

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

\(\Leftrightarrow x^4+2x^2+1-x^2+4x-4=0\)

\(\Leftrightarrow\left(x^2+1\right)^2-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x^2+1-x+2\right)\left(x^2+1+x-2\right)=0\)

\(\Leftrightarrow\left(x^2-x+3\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow x^2+x-1=0\)

\(\Leftrightarrow x=\dfrac{-1\pm\sqrt{5}}{2}\)

Cách khác: (Tổng quát)

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

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

\(\Leftrightarrow\left(x^2+y\right)^2-\left(2y-1\right)\left[x^2-\dfrac{4x}{2y-1}+\dfrac{4}{\left(2y-1\right)^2}\right]+\dfrac{4}{2y-1}-y^2-3=0\)

\(\Leftrightarrow\left(x^2+y\right)^2-\left(2y-1\right)\left(x-\dfrac{2}{2y-1}\right)^2+\dfrac{4}{2y-1}-y^2-3=0\left(1\right)\)

Ta mong muốn: \(\dfrac{4}{2y-1}-y^2-3=0\)

\(\Leftrightarrow2y^3-y^2-6y+7=0\)

\(\Leftrightarrow y=1\)

Khi đó: 

\(\left(1\right)\Leftrightarrow\left(x^2+y\right)^2-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x^2-x+3\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow x^2+x-1=0\)

\(\Leftrightarrow x=\dfrac{-1\pm\sqrt{5}}{2}\)

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

\(\Leftrightarrow x^4+2x^2+1-x^2+4x-4=0\)

\(\Leftrightarrow\left(x^2+1\right)^2-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x^2+1-x+2\right)\left(x^2+1+x-2\right)=0\)

\(\Leftrightarrow x^2+x-1=0\)

\(\text{Δ}=1^2-4\cdot1\cdot\left(-1\right)=5\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là: 

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{5}}{2}\\x_2=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)

\(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)-3x^2=0\)

\(\Leftrightarrow\left[\left(x+2\right)\left(x+12\right)\right]\left[\left(x+3\right)\left(x+8\right)\right]-3x^2=0\)

\(\Leftrightarrow\left(x^2+14x+24\right)\left(x^2+11x+24\right)-3x^2=0\)

Đặt \(x^2+11x+24=a\)

\(\Rightarrow pt:a\left(a+3x\right)-3x^2=0\)

\(\Leftrightarrow a^2+3ax-3x^2=0\)

\(\Leftrightarrow4a^2+12ax-12x^2=0\)

\(\Leftrightarrow\left(2a+3x\right)^2=21x^2\)

\(\Leftrightarrow\orbr{\begin{cases}2a+3x=x\sqrt{21}\\2a+3x=-x\sqrt{21}\end{cases}}\)

*Với \(2a+3x=x\sqrt{21}\)

\(\Leftrightarrow2x^2+22x+48+3x-x\sqrt{21}=0\)

\(\Leftrightarrow2x^2+x\left(25-\sqrt{21}\right)+48=0\)

Có \(\Delta=262-50\sqrt{21}>0\)

Nên pt có nghiệm \(x=\frac{\sqrt{21}-25\pm\sqrt{262-50\sqrt{21}}}{4}\)

Trường hợp còn lại làm tương tự

x^3+5x^2-11=0

=>\(x\in\left\{-4,44;-1,88;1,32\right\}\)

\(ĐK:-5\le x\le3\)

Đặt \(\sqrt{x+5}+\sqrt{3-x}=t\ge0\Leftrightarrow t^2-8=2\sqrt{15-2x-x^2}\), PTTT:

\(t-t^2+8-2=0\\ \Leftrightarrow t^2-t-6=0\\ \Leftrightarrow t=3\left(t\ge0\right)\\ \Leftrightarrow2\sqrt{15-2x-x^2}=3^2-8=1\\ \Leftrightarrow60-8x-4x^2=1\\ \Leftrightarrow4x^2+8x-59=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+3\sqrt{7}}{2}\left(tm\right)\\x=\dfrac{-2-3\sqrt{7}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy nghiệm pt là ...