ĐKXĐ: z>0

pt<=> \(\frac{x^3+3x^2\sqrt[3]{3x-2}-12x+\sqrt{x}-\sqrt{x}-8}{x}=0\)

<=> \(x^3+3x^2\sqrt[3]{3x+2}-12x-8=0\)

<=> \(3x^2\sqrt[3]{3x-2}-6x^2+x^3-6x^2+12x-8=0\)

<=> \(3x^2\left(\sqrt[3]{3x-2}-2\right)+\left(x-2\right)^3=0\)

<=> \(3x^2\cdot\frac{3x-2-8}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^3=0\)

<=> \(\left(x-2\right)\left(\frac{9x^2}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^2\right)=0\)

<=> \(x=2\)( vì cái trong ngoặc thứ 2 luôn dương vs mọi x>0)

vậy x=2

Một bài làm rất hay !

hihi thank ad vân ..

\(\frac{x^2}{\sqrt{3x-2}}-\frac{\sqrt{\left(3x-2\right)\left(3x-2\right)}}{\sqrt{3x-2}}=1-x\Leftrightarrow\frac{x^2-3x+2}{\sqrt{3x-2}}-1+x=0\Leftrightarrow x^2-3x+2-\sqrt{3x-2}+x\sqrt{3x-2}=0\Leftrightarrow\left(x-2\right)\left(x-1\right)+\sqrt{3x-2}\left(x-1\right)=\left(x-1\right)\left(x-2+\sqrt{3x-2}\right)\Leftrightarrow\hept{\begin{cases}x-1=0\\x-2+\sqrt{3x-2}=0\end{cases}\Leftrightarrow}x=1\)

ĐIều kiện x >2/3

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

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

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

\(\Leftrightarrow\left(x-\sqrt{3x-2}\right)^2=0\)

\(\Leftrightarrow x-\sqrt{3x-2}=0\Leftrightarrow x=\sqrt{3x-2}\)

vì ta bình phương 2 vế ta có:

x² = 3x-2

,<=> x²-3x+2 = 0

ta có x₁= 1 (thỏa mãn) ; x₂ = 2 (thỏa mãn)

Vậy:......................................

Áp dụng bđt Côsi

\(\frac{x}{\sqrt{3x-2}}+\frac{\sqrt{3x-2}}{x}=2\left(đk:x>\frac{2}{3}\right)\)

Sử dụng bất đẳng thức AM-GM ta có : 

\(\frac{x}{\sqrt{3x-2}}+\frac{\sqrt{3x-2}}{x}\ge2\sqrt{\frac{x\sqrt{3x-2}}{\sqrt{3x-2}x}}=2\)

Đẳng thức xảy ra khi và chỉ khi \(\frac{x}{\sqrt{3x-2}}=\frac{\sqrt{3x-2}}{x}\)

\(< =>x^2=3x-2< =>x^2-3x+2=0\)

Ta dễ thấy \(a+b+c=1-3+2=0\)

Nên phương trình trên sẽ có nghiệm là \(\left\{1;2\right\}\)

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

Đk:\(\sqrt{3x-2}\ge0\Rightarrow3x-2\ge0\Rightarrow x\ge\frac{2}{3}\)

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

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

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

\(\Leftrightarrow x^2+3x-2=2x\sqrt{3x-2}\)

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

\(\Leftrightarrow x^4+6x^3+5x^2-12x+4=4x^2\left(3x-2\right)\)

\(\Leftrightarrow x^4+6x^3+5x^2-12x+4=12x^3-8x^2\)

\(\Leftrightarrow x^4-6x^3+13x^2-12x+4=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-2\right)^2=0\\\left(x-1\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=1\end{array}\right.\)(thỏa mãn)

Vậy pt có nghiệm là \(\left[\begin{array}{nghiempt}x=2\\x=1\end{array}\right.\)

ĐKXĐ : \(x\ne-\frac{1}{3}\)

Ta có : \(\sqrt{x^2+x+2}=\frac{3x^2+3x+2}{3x+1}\)

\(\Leftrightarrow\sqrt{x^2+x+2}-2=\frac{3x^2+3x+2}{3x+1}-2\)

\(\Leftrightarrow\frac{x^2+x+2-4}{\sqrt{x^2+x+2}+2}=\frac{3x^2+3x+2-6x-2}{3x+1}\)

\(\Leftrightarrow\frac{x^2+x-2}{\sqrt{x^2+x+2}+2}=\frac{3x^2-3x}{3x+1}\)

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

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

\(\Leftrightarrow x=1\)( Thỏa mãn )