Theo đề: \(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)

\(\Rightarrow\sqrt[3]{x^3+5x^2}=1+\sqrt{\frac{5x^2-2}{6}}\)

\(Đkxđ:x^2\ge\frac{2}{5}\)

Đặt: \(\hept{\begin{cases}\sqrt[3]{x^3+5x^2}=u\\\sqrt{\frac{5x^2-2}{6}}=v\ge0\end{cases}}\)

Ta được: \(\hept{\begin{cases}x^3+5x^2=u^3\\5x^2-2=6v^2\Rightarrow x^3+2=\left(v-1\right)^3+2\Leftrightarrow x=v-1\\u=1+v\end{cases}}\)

Từ trên ta giải được nghiệm: \(x=-6+2\sqrt{7}\)

\(ĐKXĐ:x\le3\)

\(\Leftrightarrow\frac{5x+2\sqrt{3-x}-x}{4}>\frac{6-4+3\sqrt{3-x}}{6}\Leftrightarrow\frac{6x+3\sqrt{3-x}}{6}-\frac{2+3\sqrt{3-x}}{6}>0\Leftrightarrow3x-1>0\Leftrightarrow x>\frac{1}{3}\)

Vậy \(\frac{1}{3}

ĐKXĐ:

\(\left(2x+2-2\sqrt{5x-1}\right)+\left(\sqrt{5x^2+x+3}-\left(2x+1\right)\right)+x^2-3x+2=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{x+1+\sqrt{5x-1}}+\dfrac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+x^2-3x+2=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{2}{x+1+\sqrt{5x-1}}+\dfrac{1}{\sqrt{5x^2+x+3}+2x+1}+1\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\)

a)x=6

b)x=6

d)x=0.2