Question

Giải hệ\(\left\{{}\begin{matrix}7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\\\sqrt[3]{4x+y+1}+\sqrt{3x+2y}=4\end{matrix}\right.\)

Answer 1

Tham Khảo:

https://olm.vn/hoi-dap/detail/264041645597.html

Sai thì hong bít j đâu ;-;

Answer 2

\(7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\)

\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=0\)

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

\(\Leftrightarrow2x-1=x-y\)

\(\Leftrightarrow y=1-x\)

Thế xuống dưới:

\(\sqrt[3]{3x+2}+\sqrt{x+2}=4\)

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

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{\sqrt[3]{\left(3x+2\right)^2}+2\sqrt[3]{3x+2}+4}+\dfrac{1}{\sqrt{x+2}+2}\right)=0\)