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

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 ..

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 \(\left\{1;2\right\}\)

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

\(\Leftrightarrow5x^3+3x^2+3x-2=\dfrac{x^4}{4}+x^2\left(3x-\dfrac{1}{2}\right)+\left(3x-\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow5x^3+3x^2+3x-2=\dfrac{x^4}{4}+3x^3-\dfrac{x^2}{2}+9x^2-3x+\dfrac{1}{4}\)

\(\Leftrightarrow20x^3+12x^2+12x-8=x^4+12x^3-2x^2+36x^2-12x+1\)

\(\Leftrightarrow x^4-8x^3+22x^2-24x+9=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(7x^3-7x^2\right)+\left(15x^2-15x\right)-\left(9x-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+15x-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-x^2\right)-\left(6x^2-6x\right)+\left(9x-9\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x^2-6x+9\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy pt có nghiệm \(x=\left\{1;3\right\}\)