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

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: 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 ..

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

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

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

\(\hept{\begin{cases}\frac{1}{3x}+\frac{2x}{3y}=\frac{x+\sqrt{y}}{2x^2+y}\left(1\right)\\\sqrt{y+\sqrt{y}+x+2}+\sqrt{3x+1}=5\left(2\right)\end{cases}}\)

\(ĐK:y>0;\frac{-1}{3}\le x\ne0;y+\sqrt{y}+x+2\ge0\)

Đặt \(\sqrt{y}=tx\Rightarrow y=t^2x^2\)thay vào (1), ta được: \(\frac{1}{3x}+\frac{2x}{3t^2x^2}=\frac{x+tx}{2x^2+t^2x^2}\)

Rút gọn biến x ta đưa về phương trình ẩn t : \(\left(t-2\right)^2\left(t^2+t+1\right)=0\Leftrightarrow t=2\Leftrightarrow\sqrt{y}=2x\ge0\)

Thay vào (2), ta được: \(\sqrt{4x^2+3x+2}+\sqrt{3x+1}=5\)\(\Leftrightarrow\left(\sqrt{4x^2+3x+2}-3\right)+\left(\sqrt{3x+1}-2\right)=0\)\(\Leftrightarrow\frac{\left(x-1\right)\left(4x+7\right)}{\sqrt{4x^2+3x+2}+3}+\frac{3\left(x-1\right)}{\sqrt{3x+1}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{4x+7}{\sqrt{4x^2+3x+2}+3}+\frac{3}{\sqrt{3x+1}+2}\right)=0\)

Dễ thấy \(\frac{4x+7}{\sqrt{4x^2+3x+2}+3}+\frac{3}{\sqrt{3x+1}+2}>0\)nên \(x-1=0\Leftrightarrow x=1\Rightarrow y=4\)

Vậy hệ phương trình có 1 nghiệm duy nhất \(\left(x,y\right)=\left(1,4\right)\)

quy đồng lên bạn ơi , rồi đặt ẩn

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

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

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

\(\Leftrightarrow x^2-4x+4=3x-2\Leftrightarrow x=6;x=1\left(\text{nhận cả 2}\right)\)

Vậy................

mình ko biết