Question

[chuyên toán thcs ( Cool Team )]

Giải hpt 

\(\hept{\begin{cases}x^2+y^2+xy+1=4y\\\left(x^2+1\right)\left(x+y-2\right)=y\end{cases}}\)

Answer 1

[Nyatmax]

HPT

\(\Leftrightarrow\hept{\begin{cases}\left(x^2+1\right)+y\left(x+y-2\right)=2y\\\left(x^2+1\right)\left(x+y-2\right)=y\end{cases}}\)

y=0 khong phai nghiem cua hpt

\(\Rightarrow\hept{\begin{cases}\left(\frac{x^2}{y}+\frac{1}{y}\right)+\left(x+y-2\right)=2\\\left(\frac{x^2}{y}+\frac{1}{y}\right)\left(x+y-2\right)=1\end{cases}}\)

Dat \(\hept{\begin{cases}\frac{x^2}{y}+\frac{1}{y}=a\\x+y-2=b\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a+b=2\\ab=1\end{cases}}\)

Đến đây là ngon

Answer 2

[chuyên toán thcs ( Cool Team )]

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

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

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

\(\Leftrightarrow x^3=6+3\sqrt[2]{9-8}.x\)

\(\Leftrightarrow x^3=6+3x\)