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a) Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=a\\\frac{1}{y-1}=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}5a+b=10\\a-3b=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15a+3b=30\\a-3b=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-3b=18\\16a=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=3\\\frac{1}{y-1}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{4}{5}\end{matrix}\right.\)
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b) Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=a\\\frac{1}{\sqrt{y+6}}=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\5a+3b=\frac{13}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\51a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{306}\\b=\frac{-43}{612}\end{matrix}\right.\)( loại vì \(a,b>0\) )
Vậy hệ vô nghiệm
Is that true .-.
Cho xin solve lại câu b)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}21a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5a+3b=\frac{13}{6}\\41a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{246}\\b=\frac{8}{123}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=\frac{97}{246}\\\frac{1}{\sqrt{y+6}}=\frac{8}{123}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{126379}{9409}\\y=\frac{14745}{64}\end{matrix}\right.\)
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a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
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b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
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c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)
\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)
Vậy hệ pt vô nghiệm
d) Nhân 3 pt đầu rồi thu gọn
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 \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-2}}=u>0\\\frac{1}{x+y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5u-2v=4\\4u-3v=\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{\sqrt{x-2}}=1\\\frac{1}{x+y}=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\x+y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
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: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{4x+y}=a\ge0\\\sqrt{x+2y}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{2a^2-b^2}{7}\\y=\frac{4b^2-a^2}{7}\end{matrix}\right.\)
Ta được: \(\left\{{}\begin{matrix}a+b=5\\\frac{5\left(2a^2-b^2\right)}{21}-\frac{4b^2-a^2}{42}+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\21a^2-14b^2+42b-84=0\end{matrix}\right.\)
\(\Rightarrow21\left(5-b\right)^2-14b^2+42b-84=0\)
\(\Leftrightarrow b^2-24b+63=0\Rightarrow\left[{}\begin{matrix}b=21\Rightarrow a=-16\left(l\right)\\b=3\Rightarrow a=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{4x+y}=2\\\sqrt{x+2y}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x+y=4\\x+2y=9\end{matrix}\right.\) \(\Leftrightarrow...\)
Lấy PT1 trừ PT2 ta được
\(\sqrt{4x+y}-\frac{5}{3}x+\frac{1}{6}y=3\)
\(\Leftrightarrow6\sqrt{4x+y}-10x+y=18\)
đặt \(\sqrt{4x+y}=a\left(a\ge0\right)\)
\(\Rightarrow6a-\frac{5a^2-7y}{2}=18\)
\(\Leftrightarrow12a-5a^2+7y=36\)
Giải a theo y, rồi thay vào
ĐKXĐ : \(\left\{{}\begin{matrix}x>7\\y>-6\end{matrix}\right.\)
- Đặt \(\frac{1}{\sqrt{x-7}}=a,\frac{1}{\sqrt{y+6}}=b\) ( \(a,b\ne0\) ) vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}7a-4b=\frac{5}{3}\\5a+3b=\frac{13}{6}\end{matrix}\right.\)
( đoạn này ruễ tự giải nhoa )
=> \(\left\{{}\begin{matrix}a=\frac{1}{3}\\b=\frac{1}{6}\end{matrix}\right.\)( TM )
- Thay lại \(\frac{1}{\sqrt{x-7}}=a,\frac{1}{\sqrt{y+6}}=b\) vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=\frac{1}{3}\\\frac{1}{\sqrt{y+6}}=\frac{1}{6}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-7=9\\y+6=36\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=16\\y=30\end{matrix}\right.\) ( TM )
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THẠNKS