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mình giải tắt nhé vì mình không giỏi dùng công thức. Thông cảm nha.
1.
\(\left\{{}\begin{matrix}3x-y=2m+3\\x+y=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m}{4}+1\\y=\dfrac{-5m}{4}\end{matrix}\right.\)
vậy phương trình có nghiệm duy nhất là \(\left(\dfrac{m}{4}+1;\dfrac{-5m}{4}\right)\)
Thay vào đẳng thức ta được:
\(\left(\dfrac{m}{4}+1\right)^2+\left(\dfrac{-5m}{4}\right)^2=5\\ \Leftrightarrow x=\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x-6y=66\\-3x=-9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x^2+xy+3=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+x\left(1-x\right)+3=0\)
\(\Leftrightarrow x+3=0\Rightarrow x=-3\Rightarrow y=4\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{2x-5}{3}\\x^2-y^2=40\end{matrix}\right.\)
\(\Rightarrow x^2-\left(\frac{2x-5}{3}\right)^2-40=0\)
\(\Leftrightarrow9x^2-\left(4x^2-20x+25\right)-360=0\)
\(\Leftrightarrow5x^2+20x-385=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\Rightarrow y=3\\x=-11\Rightarrow y=-9\end{matrix}\right.\)
d.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{36-3x}{2}\\\left(x-2\right)\left(y-3\right)=18\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)\left(\frac{36-3x}{2}-3\right)=18\)
\(\Leftrightarrow\left(x-2\right)\left(10-x\right)=12\)
\(\Leftrightarrow-x^2+12x-32=0\Rightarrow\left[{}\begin{matrix}x=4\Rightarrow y=12\\x=8\Rightarrow y=6\end{matrix}\right.\)
1)
\(\left\{{}\begin{matrix}x+y=4\\2x+3y=m\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+3y=12\\2x+3y=m\end{matrix}\right.\)
trừ 2 vế của pt cho nhau ta tìm được
\(\left\{{}\begin{matrix}x=12-m\\y=m-8\end{matrix}\right.\)
để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m< 12\\m< 8\end{matrix}\right.\Rightarrow}m< 8}\)
a/ Bạn tự giải
b/ ĐKXĐ:...
Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)
Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)
\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)
\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)
Chắc bạn ghi sai đề, nghiệm quá xấu
3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)
4/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)
\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)
Còn lại tương tự
Vì \(\dfrac{1}{2}\ne\dfrac{-2}{3}\)
nên hệ luôn có nghiệm duy nhất
a: \(\left\{{}\begin{matrix}x-2y=-3m-4\\2x+3y=8m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-4y=-6m-8\\2x+3y=8m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-4y-2x-3y=-6m-8-8m+1\\2x+3y=8m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-7y=-14m-7\\2x=8m-1-3y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2m+1\\2x=8m-1-6m-3=2m-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2m+1\\x=m-2\end{matrix}\right.\)
Đặt \(A=y^2+3x-1\)
\(=\left(2m+1\right)^2+3\left(m-2\right)-1\)
\(=4m^2+4m+1+3m-6-1\)
\(=4m^2+7m-6\)
\(=4\left(m^2+\dfrac{7}{4}m-\dfrac{3}{2}\right)\)
\(=4\left(m^2+2\cdot m\cdot\dfrac{7}{8}+\dfrac{49}{64}-\dfrac{145}{64}\right)\)
\(=4\left(m+\dfrac{7}{8}\right)^2-\dfrac{145}{16}>=-\dfrac{145}{16}\)
Dấu '=' xảy ra khi m=-7/8
b: Đặt B=x^2-y^2
\(=\left(m-2\right)^2-\left(2m+1\right)^2\)
\(=m^2-4m+4-4m^2-4m-1\)
\(=-3m^2-8m+3\)
\(=-3\left(m^2+\dfrac{8}{3}m-1\right)\)
\(=-3\left(m^2+2\cdot m\cdot\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{25}{9}\right)\)
\(=-3\left(m+\dfrac{4}{3}\right)^2+\dfrac{25}{3}< =\dfrac{25}{3}\)
Dấu '=' xảy ra khi m=-4/3