\(x^2y+xy^2+x+y=117\)
\(\Leftrightarrow12x+12y+x+y=117\)
\(\Leftrightarrow13\left(x+y\right)=117\)
\(\Leftrightarrow x+y=9\)
\(\Rightarrow\left(x+y\right)^3=9^3\)
\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=729\)
\(\Leftrightarrow x^3+y^3+3.12.9=729\)
\(\Rightarrow x^3+y^3=729-324=405\)
\(x^2y+xy^2+x+y=117\\ \Leftrightarrow xy\left(x+y\right)+x+y=117\\ \Leftrightarrow\left(xy+1\right)\left(x+y\right)=117\\ \Leftrightarrow x+y=\frac{117}{12+1}=9\)
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\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left(x-y\right)^2+xy=\left(x+y\right)^3+xy=9^3+12=741\)
sorry sai ngay khúc cuối :)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left[\left(x-y\right)^2+xy\right]=9.\left[9^2+12\right]=837\)
Ta có:
\(x^2y+xy^2+x+y=117.\)
\(\Leftrightarrow\left(x^2y+xy^2\right)+\left(x+y\right)=117\)
\(\Leftrightarrow xy.\left(x+y\right)+\left(x+y\right)=117\)
\(\Leftrightarrow\left(x+y\right).\left(xy+1\right)=117\)
Mà \(x.y=12\)
\(\Leftrightarrow xy+1=12+1\)
\(\Leftrightarrow xy+1=13\)
\(\Leftrightarrow\left(x+y\right).13=117\)
\(\Leftrightarrow\left(x+y\right)=117:13\)
\(\Rightarrow x+y=9\)
Lại có: \(x^3+y^3\)
\(=\left(x+y\right)^3-3xy.\left(x+y\right)\)
\(=9^3-3.12.9\)
\(=729-324\)
\(=405.\)
Vậy \(x^3+y^3=405.\)
Chúc bạn học tốt!
