Question

giải pt

`a,(x+\sqrt{3})+4(x^2-3)=0`

Answer 1

`a,(x+\sqrt{3})+4(x^2-3)=0`

`<=>(x+\sqrt{3})+4(x-\sqrt{3})(x+\sqrt{3})=0`

`<=>(x+\sqrt{3})[4(x-\sqrt{3}+1]=0`

`<=>(x+\sqrt{3})(4x-4\sqrt{3}+1)=0`

`<=>` \(\left[ \begin{array}{l}x+\sqrt{3}=0\\4x-4\sqrt{3}+1=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\sqrt{3}\\4x=4\sqrt{3}-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\sqrt{3}\\x=\sqrt{3}-\dfrac{1}{4}\end{array} \right.\) 

Vậy phương trình có tập nghiệm `S={-\sqrt{3},\sqrt{3}-1/4}`

Answer 2

\(\Leftrightarrow\left(x+\sqrt{3}\right)+4\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{3}\right)\left(1+4x-4\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\dfrac{4\sqrt{3}-1}{4}\end{matrix}\right.\)

Answer 3

a) Ta có: \(\left(x+\sqrt{3}\right)+4\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{3}\right)+4\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{3}\right)\left(1+4x-4\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=0\\4x+1-4\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\4x=4\sqrt{3}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=\dfrac{4\sqrt{3}-1}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{3};\dfrac{4\sqrt{3}-1}{4}\right\}\)