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TG
1
BT
2 tháng 12 2017
Điều kiện \(x\ne\pm3;y\ne-2\):
\(P=\frac{2x+3y}{xy+2x-3y-6}-\frac{6-xy}{xy+2x+3y+6}-\frac{x^2+9}{x^2-9}.\)
=> \(P=\frac{2x+3y}{\left(y+2\right)\left(x-3\right)}-\frac{6-xy}{\left(y+2\right)\left(x+3\right)}-\frac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)
\(P=\frac{\left(2x+3y\right)\left(x+3\right)-\left(6-xy\right)\left(x-3\right)-\left(x^2+9\right)\left(y+2\right)}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)
\(P=\frac{2x^2+3xy+6x+9y-6x+x^2y+18-3xy-x^2y-9y-2x^2-18}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)
\(P=\frac{0}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}=0\)
=> P=0 (với mọi x khác 3, -3 và y khác -2)
PH
1
H9
HT.Phong (9A5)
CTVHS
3 tháng 7 2023
a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
CD
0
CD
0
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
\(2x-5\sqrt{xy}+3y\\ =2x-2\sqrt{xy}-3\sqrt{xy}+3y\\ =2\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\\ =\left(2\sqrt{x}-3\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(2x-5\sqrt{xy}+3y\)
\(=2x-2\sqrt{xy}-3\sqrt{xy}+3y\)
\(=2\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(2\sqrt{x}-3\sqrt{y}\right)\)