\(\Leftrightarrow3\sqrt{3}x^3+9x^2+3\sqrt{3}x-9=0\)

\(\Leftrightarrow\left(\sqrt{3}x\right)^3+3.\left(\sqrt{3}x\right)^2.1+3.\left(\sqrt{3}x\right).1^2+1^3-10=0\)

\(\Leftrightarrow\left(\sqrt{3}x+1\right)^3=10\)

\(\Leftrightarrow\sqrt{3}x+1=\sqrt[3]{10}\)

\(\Leftrightarrow x=\frac{\sqrt[3]{10}-1}{\sqrt{3}}\)

`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}`

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

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

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

\(\sqrt[3]{3-x}+\sqrt[3]{x-1}=0\)

\(\Leftrightarrow\sqrt[3]{3-x}=-\sqrt[3]{x-1}\)

\(\Leftrightarrow3-x=1-x\)

\(\Leftrightarrow0x=2\left(voli\right)\)

Vậy \(ptvn\)

_dsaddaadad

a) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)

\(\Leftrightarrow7\left|x-1\right|=35\)

\(\Leftrightarrow\left|x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)

c) ĐKXĐ: \(x\ge0\)

Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)

\(\Leftrightarrow x-1=x+\sqrt{x}-6\)

\(\Leftrightarrow\sqrt{x}-6=-1\)

\(\Leftrightarrow\sqrt{x}=5\)

hay x=25(nhận)

 Em cảm ơn ạ ❤️❤️❤️

a) Đk: \(\hept{\begin{cases}x^2-4x+1\ge0\\x+1\ge0\end{cases}}\)

\(\sqrt{x^2-4x+1}=\sqrt{x+1}\)

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

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

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)thỏa mãn điều kiện

Vậy x=0 hoặc x=5

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

Đk: x>=3 hoặc x=1

pt  (1)<=> \(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)

<=> \(\sqrt{x-1}=0\)(vì\(\sqrt{x-3}+1>0\)mọi x )

<=> x-1=0

<=> x=1 ( thỏa mãn điều kiện)