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\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)
Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)
Làm nốt nha
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:...
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)
\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)
a/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=u\\\frac{1}{2y+1}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u+v=\frac{6}{5}\\3u-2v=\frac{11}{10}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\frac{1}{2}\\v=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=2\\2y+1=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
b/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}x+y=u\\\sqrt{x+1}=v\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u+v=4\\u-3v=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=1\\\sqrt{x+1}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x+1=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
ĐKXĐ : \(\left\{{}\begin{matrix}x+2\ge0\\2x-y\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-2\\2x\ne y\end{matrix}\right.\)
Ta có : \(\left\{{}\begin{matrix}\frac{\sqrt{x+2}}{3}+\frac{1}{2x-y}=\frac{4}{3}\\2\sqrt{x+2}-\frac{3}{y-2x}=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{\sqrt{x+2}}{3}+\frac{1}{2x-y}=\frac{4}{3}\\2\sqrt{x+2}+\frac{3}{2x-y}=5\end{matrix}\right.\)
- Đặt \(a=\sqrt{x+2},b=\frac{1}{2x-y}\) ( \(a\ge0,\frac{1}{b}\ne0\) ) ta được hệ :
\(\left\{{}\begin{matrix}\frac{a}{3}+b=\frac{4}{3}\\2a+3b=5\end{matrix}\right.\)
( Đoạn này bấm máy cho nhanh nha )
=> \(\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) ( TM )
- Thay lại \(a=\sqrt{x+2},b=\frac{1}{2x-y}\) ta được :
\(\left\{{}\begin{matrix}\sqrt{x+2}=1\\\frac{1}{2x-y}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+2=1\\2x-y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-1\\2\left(-1\right)-y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\) ( TM )
Vậy ...
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{\sqrt{x-2}}-\frac{2}{x+y}=4\\\frac{4}{\sqrt{x-2}}-\frac{3}{x+y}+1=\frac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{\sqrt{x-2}}-\frac{2}{x+y}=4\\\frac{4}{\sqrt{x-2}}-\frac{3}{x+y}=\frac{5}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-2}}=u>0\\\frac{1}{x+y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5u-2v=4\\4u-3v=\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{\sqrt{x-2}}=1\\\frac{1}{x+y}=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\x+y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)