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!

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

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

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

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

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

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

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

hình như em ghi sai đề rồi em nhé vì câu a không cũng 1 dạng sẽ không đưa về hằng đẳng thức được!

\(ĐK:-5\le x\le3\)

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 ...

\(2x^4+7x^3+x^2-7x-3=0\)

\(\Leftrightarrow2x^4+7x^3+3x^2-2x^2-7x-3=0\)

\(\Leftrightarrow\left(2x^4+7x^3+3x^2\right)-\left(2x^2+7x+3\right)=0\)

\(\Leftrightarrow x^2\left(2x^2+7x+3\right)-\left(2x^2+7x+3\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2+7x+3\right)=0\)

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

\(\Leftrightarrow x\in\left\{\pm1;\frac{-1}{2};-3\right\}\)

https://olm.vn/hoi-dap/question/413076.html

\(3x^4+2x^3-10x^2+2x+3=0\)

\(\Leftrightarrow3x^4-6x^3+3x^2+8x^3-16x^2+8x+3x^2-6x+3=0\)

\(\Leftrightarrow3x^2\left(x^2-2x+1\right)+8x\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)\left(3x^2+8x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(3x^2+8x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(3\left(x+\dfrac{4}{3}\right)^2-\dfrac{7}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3\left(x+\dfrac{4}{3}\right)^2-\dfrac{7}{3}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8\pm\sqrt{28}}{6}\end{matrix}\right.\)