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!

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

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

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

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

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

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

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

\(2x^3-7x^2+4x+1=0\)

\(\Leftrightarrow2x^2\left(x-1\right)-5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^2-5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x^2-5x-1=0\end{cases}}\) Đến đây tự làm tiếp nha

(x^2+2x+1)(x+2)-x^2(x-3)-7x(x-1)=3x-9

<=>x³+4x²+5x+2-x³+3x²-7x²+7x=3x-9

<=>14x+2=3x-9

<=>14x-3x=-9-2

<=>11x=-11

<=>x=-1

vậy S={-1}

trieu dang làm đúng rùi. ****

ĐKXĐ: \(x\ge\dfrac{2}{7}\)

\(\sqrt{5x^2-5x+3}-\left(x+1\right)+2x-\sqrt{7x-2}+4x^2-7x+2=0\)

\(\Leftrightarrow\dfrac{4x^2-7x+2}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{4x^2-7x+2}{2x+\sqrt{7x-2}}+4x^2-7x+2=0\)

\(\Leftrightarrow\left(4x^2-7x+2\right)\left(\dfrac{1}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{1}{2x+\sqrt{7x-2}}+1\right)=0\)

Ta có \(\dfrac{1}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{1}{2x+\sqrt{7x-2}}+1>0\)

\(\Rightarrow4x^2-7x+2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7-\sqrt{17}}{8}\\x=\dfrac{7+\sqrt{17}}{8}\end{matrix}\right.\)

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