ĐKXĐ: x>=-1

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

=>\(4x^2-2\sqrt{x+1}-x-2=0\)

=>\(4x^2+3x-4x-3+1-2\sqrt{x+1}=0\)

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

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

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

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

=>4x+3=0

=>x=-3/4(nhận)

Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$

PT $\Leftrightarrow \frac{x}{\sqrt{x+2}}=\sqrt{3x+1}-\sqrt{x+1}$

$\Leftrightarrow \frac{x}{\sqrt{x+2}}=\frac{2x}{\sqrt{3x+1}+\sqrt{x+1}}$

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

Xét các TH:

TH1: $x=0$ (thỏa mãn)

TH2: $\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}$

$\Leftrightarrow \sqrt{3x+1}+\sqrt{x+1}=2\sqrt{x+2}$

$\Rightarrow 4x+2+2\sqrt{(3x+1)(x+1)}=4(x+2)$

$\Leftrightarrow \sqrt{(3x+1)(x+1)}=3$

$\Rightarrow (3x+1)(x+1)=9$

$\Leftrightarrow 3x^2+4x-8=0$

$\Rightarrow x=\frac{-2\pm 2\sqrt{7}}{3}$

Kết hợp với ĐKXĐ suy ra $x=\frac{-2+2\sqrt{7}}{3}$

Vậy............

ĐKXĐ: ...

\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)

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

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

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

\(\Leftrightarrow...\)

ĐKXĐ: x \(\ge\)\(\dfrac{1}{3}\)

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

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

  \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{3x-1}\\1=\sqrt{3x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)(t/m x \(\ge\)\(\dfrac{1}{3}\))

Vậy.....................

\(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)(Đk x≥\(\dfrac{1}{3}\))

ta có:\(x\left(3-\sqrt{3x-1}\right)\)

=\(3x-x\sqrt{3x-1}\)

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

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

Ta có \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

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

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

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

ta có \(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

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

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

⇔​​\(\sqrt{3x-1}=\sqrt{x+1}\)

⇔​\(3x-1=x+1\)

⇔\(2x=2\)

⇔x=1(N)

​Vậy x=1

Hệ \(\Leftrightarrow x+1+3x-1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{3x-1}\right)=x-1\)

\(\Leftrightarrow3x+1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow3x+1=-3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}\)

\(\Leftrightarrow27x^3+9x+27x^2+1=-27\left(x^2-1\right)\left(3x-1\right)\)

\(\Leftrightarrow27x^3+9x+27x^2+1+81x^3-81x-27x^2+27=0\)

\(\Leftrightarrow108x^3-72x+28=0\)

\(\Leftrightarrow x^3-\dfrac{2}{3}x+\dfrac{7}{27}=0\)

- AD công thức các đa nô :

\(\Rightarrow x=\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}+\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}+\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}-\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}\)

\(\Rightarrow x\approx-0,96685\)

Hummmm

Dạ em không biết ạ,tại vì em mới học lớp 4 ạ,em xin lỗi ạ

ĐKXĐ: \(x\ge1\)

\(\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}\)

\(\Leftrightarrow5x-1=3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}\)

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

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

\(\Leftrightarrow11x^2-24x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{11}\left(loại\right)\\x=2\end{matrix}\right.\)

\(PT\Leftrightarrow\sqrt{3x+1}=\sqrt{x+4}+1\\ \Leftrightarrow3x+1=x+5+2\sqrt{x+4}\\ \Leftrightarrow2x-4=2\sqrt{x+4}\\ \Leftrightarrow x-2=\sqrt{x+4}\\ \Leftrightarrow x^2-4x+4=x+4\\ \Leftrightarrow x^2-5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Thử lại ta thấy x=0 ko thỏa mãn

Vậy PT có nghiệm x=5

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\sqrt{3x+1}=1+\sqrt{x+4}\)

\(\Leftrightarrow3x+1=1+x+4+2\sqrt{x+2}\)

\(\Leftrightarrow x+2-\sqrt{x+2}-4=0\)

Đặt \(\sqrt{x+2}=t\ge0\)

\(\Rightarrow t^2-t-4=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{17}}{2}\\t=\dfrac{1-\sqrt{17}}{2}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+2}=\dfrac{1+\sqrt{17}}{2}\)

\(\Rightarrow x=\dfrac{5+\sqrt{17}}{2}\)

\(\sqrt{x+11}-\sqrt{10-3x}=\sqrt{1-x}\left(1\ge x\ge-11\right)\)

\(\Leftrightarrow\left(x+11\right)+\left(10-3x\right)-2\sqrt{\left(x+11\right)\left(10-3x\right)}=1-x\\ \Leftrightarrow-2x+21-2\sqrt{-3x^2-23x+110}=1-x\\ \Leftrightarrow-2\sqrt{-3x^2-23x+110}=x-20\\ \Leftrightarrow4\left(-3x^2-23x+110\right)=x^2-40x+400\\ \Leftrightarrow-12x^2-92x+440=x^2-40x+400\\ \Leftrightarrow13x^2+52x-40=0\)

\(\Delta=52^2-4\cdot\left(-40\right)\cdot13=4784>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\sqrt{299}-52}{26}\\x=\dfrac{4\sqrt{299}-52}{26}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\sqrt{299}-26}{13}\\x=\dfrac{2\sqrt{299}-26}{13}\end{matrix}\right.\)

Tick nha