Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
d) \(x^2+y^2-4x+4y=1\\ \Rightarrow\left(x-2\right)^2+\left(y+2\right)^2=8\)
\(\Rightarrow8=\left(x-2\right)^2+\left(y+2\right)^2\ge\left(x-2\right)^2\)
\(\Rightarrow\left(x-2\right)^2\le8\)
Mà \(\left(x-2\right)^2\) là SCP và là số chẵn nên \(\left(x-2\right)^2\in\left\{0;4\right\}\)
Th1: \(\left(x-2\right)^2=0\Rightarrow\left(y+2\right)^2=8\left(vôlí\right)\)
Th2: \(\left(x-2\right)^2=4\Rightarrow\left(y+2\right)^2=4\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=-2\\y+2=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-2\\y+2=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+2=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+2=2\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(0;-4\right);\left(0;0\right);\left(4;-4\right);\left(4;0\right)\right\}\)
a. 3x2 - 4y2 = 18
<=> \(\left\{{}\begin{matrix}3x^2=18+4y^2\\4y^2=-\left(3x^2-18\right)\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\sqrt{\dfrac{18+4y^2}{3}}\\y=\sqrt{\dfrac{-3x^2+18}{4}}\end{matrix}\right.\)
b, c, d tương tự nhé
b. 19x2 + 28y2 = 2001
<=> \(\left\{{}\begin{matrix}19x^2=2001-28y^2\\28y^2=2001-19x^2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\sqrt{\dfrac{2001-28y^2}{19}}\\y=\sqrt{\dfrac{2001-19x^2}{28}}\end{matrix}\right.\)
c. x2 = 2y2 - 8y + 3
<=> \(\left\{{}\begin{matrix}x=\sqrt{2y^2-8y+3}\\8y=2y^2+3-x^2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\sqrt{2y^2-8y+3}\\y=\dfrac{2y^2+3-x^2}{8}\end{matrix}\right.\)
d. x2 + y2 - 4x + 4y = 1
<=> \(\left\{{}\begin{matrix}x^2=1-y^2+4x-4y\\y^2=1-x^2+4x-4y\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\sqrt{1-y^2+4x-4y}\\y=\sqrt{1-x^2+4x-4y}\end{matrix}\right.\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
3/ \(x^2=2\left(y-2\right)^2-5\Rightarrow\left(\sqrt{2}y-2\sqrt{2}\right)^2-x^2=5\)
\(\Leftrightarrow\left(\sqrt{2}y-2\sqrt{2}+x\right)\left(\sqrt{2}y-2\sqrt{2}-x\right)=5\)
Lập bảng giải ra tiếp.
P/s: Cách này có vẽ không hay lắm thiết nghĩ dùng delta sẽ hay hơn nhưng để thử=)
4) \(x^2+x\left(y-2\right)+\left(y^2-y\right)=0\)
Pt trên có ẩn x.
\(\Delta=\left(y-2\right)^2-4\left(y^2-y\right)\ge0\)
\(\Leftrightarrow-3y^2+4\ge0\Leftrightarrow-\frac{2\sqrt{3}}{3}\le y\le\frac{2\sqrt{3}}{3}\)
Do y nguyên nên \(-1\le y\le1\).
Làm nốt