Question

[Sương Đặng]

giải pt

\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x^3+5x^2+4x+1\right)}=5x^3-3x^2+8\)

Answer 1

[Huyền]

\(pt\Leftrightarrow2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x+1\right)\left(x+1\right)^2}=\left(x+1\right)\left(5x^2-8x+8\right)\)\(\Leftrightarrow2\left(x+1\right)\sqrt{x}+\left(x+1\right)\sqrt{3\left(2x+1\right)}-\left(x+1\right)\left(5x^2-8x+8\right)=0\)\(\Leftrightarrow\left(x+1\right)\left(2\sqrt{x}+\sqrt{3\left(2x+1\right)}-5x^2+8x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\2\sqrt{x}+\sqrt{3\left(2x+1\right)}-5x^2-8+8x=0\circledast\end{matrix}\right.\)

Giải (*)\(2\sqrt{x}+\sqrt{3\left(2x+1\right)}-5x^2-8+8x=0\)

\(\Leftrightarrow2\sqrt{x}-2+\sqrt{3\left(2x+1\right)}-3=5x^2-8x+3\)

\(\Leftrightarrow\frac{4x-4}{2\sqrt{x}+2}+\frac{6x-6}{\sqrt{3\left(2x+1\right)}+3}=\left(x-1\right)\left(5x-3\right)\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{3\left(2x+1\right)}+3}-5x+3\right)=0\)

x=1

bạn giải nốt cái còn lại nhá