x^3+x^3y^3+y^3=17 .x+y+xy=5 mọi người gúp mình giải hệ với ạ 11/08/2021 Bởi Allison x^3+x^3y^3+y^3=17 .x+y+xy=5 mọi người gúp mình giải hệ với ạ
Đáp án: x=2,y=1 x=1,y=2 Giải thích các bước giải: \(\left\{ \begin{array}{l}{x^3} + {x^3}{y^3} + {y^3} = 17\\x + y + xy = 5\end{array} \right.\) Đặt \(\left\{ \begin{array}{l}x + y = a\\xy = b\end{array} \right.\) \(\begin{array}{l} \to \left\{ \begin{array}{l}{a^3} – 3ab + {b^3} = 17\\a + b = 5\end{array} \right.\\{a^3} – 3ab + {b^3} = 17 \leftrightarrow {(a + b)^3} – 3ab(a + b) – 3ab = 17\\ \leftrightarrow {5^3} – 3ab.5 – 3ab = 17\\ \leftrightarrow ab = 6\\ \to \left\{ \begin{array}{l}a + b = 5\\ab = 6\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}a = 5 – b\\b(5 – b) = 6\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}a = 5 – b\\ – {b^2} + 5b – 6 = 0\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}a = 5 – b\\\left[ \begin{array}{l}b = 3\\b = 2\end{array} \right.\end{array} \right. \leftrightarrow \left[ \begin{array}{l}a = 2,b = 3\\a = 3,b = 2\end{array} \right.\end{array}\) Xét a=2,b=3 \( \to \left\{ \begin{array}{l}x + y = 2\\xy = 3\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = 2 – y\\y(2 – y) = 3\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = 2 – y\\ – {y^2} + 2y = 3(VN)\end{array} \right. \to x,y \in \phi \) Xét a=3,b=2 \( \to \left\{ \begin{array}{l}x + y = 3\\xy = 2\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = 3 – y\\y(3 – y) = 2\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}x = 3 – y\\ – {y^2} + 3y = 2(VN)\end{array} \right. \to \left\{ \begin{array}{l}x = 3 – y\\\left[ \begin{array}{l}y = 1\\y = 2\end{array} \right.\end{array} \right. \leftrightarrow \left[ \begin{array}{l}x = 2,y = 1\\x = 1,y = 2\end{array} \right.\) Bình luận
Đáp án:
x=2,y=1
x=1,y=2
Giải thích các bước giải:
\(\left\{ \begin{array}{l}
{x^3} + {x^3}{y^3} + {y^3} = 17\\
x + y + xy = 5
\end{array} \right.\)
Đặt \(\left\{ \begin{array}{l}
x + y = a\\
xy = b
\end{array} \right.\)
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
{a^3} – 3ab + {b^3} = 17\\
a + b = 5
\end{array} \right.\\
{a^3} – 3ab + {b^3} = 17 \leftrightarrow {(a + b)^3} – 3ab(a + b) – 3ab = 17\\
\leftrightarrow {5^3} – 3ab.5 – 3ab = 17\\
\leftrightarrow ab = 6\\
\to \left\{ \begin{array}{l}
a + b = 5\\
ab = 6
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 5 – b\\
b(5 – b) = 6
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 5 – b\\
– {b^2} + 5b – 6 = 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 5 – b\\
\left[ \begin{array}{l}
b = 3\\
b = 2
\end{array} \right.
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
a = 2,b = 3\\
a = 3,b = 2
\end{array} \right.
\end{array}\)
Xét a=2,b=3
\( \to \left\{ \begin{array}{l}
x + y = 2\\
xy = 3
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 2 – y\\
y(2 – y) = 3
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 2 – y\\
– {y^2} + 2y = 3(VN)
\end{array} \right. \to x,y \in \phi \)
Xét a=3,b=2
\( \to \left\{ \begin{array}{l}
x + y = 3\\
xy = 2
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 3 – y\\
y(3 – y) = 2
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 3 – y\\
– {y^2} + 3y = 2(VN)
\end{array} \right. \to \left\{ \begin{array}{l}
x = 3 – y\\
\left[ \begin{array}{l}
y = 1\\
y = 2
\end{array} \right.
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
x = 2,y = 1\\
x = 1,y = 2
\end{array} \right.\)