Giải các hệ phương trình sau bằng cách đặt ẩn số phụ:
1) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}x^2+y^2=13\\3x^2-2y^2=-6\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}|x|+4|y|=18\\3|x|+|y|=10\end{matrix}\right.\)
GIẢI GIÚP MÌNH VỚI M.N
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
ba cái đồ êu!!
câu số 6 (con số của quỷ sa tăng :v)
đặt \(\left\{{}\begin{matrix}a=\left|x\right|\\b=\left|y\right|\end{matrix}\right.\) (a,b >/ 0)
hpt trở thành : \(\left\{{}\begin{matrix}a+4b=18\\3a+b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x\right|=2\\\left|y\right|=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=4\\y=-4\end{matrix}\right.\end{matrix}\right.\)
Vậy hpt có các ng (x;y) là: (có 4 nghiệm tự kết luận)
1, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\) (I) (ĐKXĐ: x, y \(\ne\)0)
Đặt \(\dfrac{1}{x}=a\) ; \(\dfrac{1}{y}=b\)
Hệ pt (I) trở thành :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\) \(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}-7b=\dfrac{-1}{3}\\a+b=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a+\dfrac{1}{21}=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\left(tm\right)\\a=\dfrac{1}{28}\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{28}\\\dfrac{1}{y}=\dfrac{1}{21}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
sorry, bận 1 chút, kết quả câu 2 là hpt có 1 nghiệm (x;y) là (1/3 ; 1/3)
2, \(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.\) (I)
(ĐKXĐ: x+2y \(\ne\)0 ; 2x+y\(\ne\)0)
Đặt \(\dfrac{1}{x+2y}=a\) ; \(\dfrac{1}{y+2x}=b\)
Hệ (I) trở thành:
\(\left\{{}\begin{matrix}2a+b=3\\4a-3b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}4a+2b=6\\4a-3b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}5b=5\\2a+b=3\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=1\\2a+1=3\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=1\left(tm\right)\\a=1\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{1}{x+2y}=1\\\dfrac{1}{y+2x}=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+2y=1\\y+2x=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+2y=1\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x+4y=2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3y=1\\x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x+2.\dfrac{1}{3}=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\y=\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)
- Thay x = \(\dfrac{1}{3}\) ; y = \(\dfrac{1}{3}\) vào x+2y \(\ne\)0 có:
\(\dfrac{1}{3}+2.\dfrac{1}{3}\ne0\) \(\Leftrightarrow\) \(1\ne0\) (luôn đúng)
- Thay x = \(\dfrac{1}{3}\) ; y = \(\dfrac{1}{3}\) vào y+2x\(\ne\)0 có:
\(\dfrac{1}{3}\) +2.\(\dfrac{1}{3}\)\(\ne\) 0 \(\Leftrightarrow\) \(1\ne0\)( luôn đúng)
Vậy hệ pt có nghiệm duy nhất \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\) (I)
(ĐKXĐ: x+1 \(\ne\)0 ; y+4\(\ne\)0)
Đặt \(\dfrac{x}{x+1}=a\) ; \(\dfrac{1}{y+4}=b\)
hệ (I) trở thành:
\(\left\{{}\begin{matrix}3a-2b=4\\2a-5b=9\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}6a-4b=8\\6a-15b=27\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}11b=-19\\3a-2b=4\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{-19}{11}\\3a-2.\dfrac{-19}{11}=4\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{-19}{11}\\a=\dfrac{2}{11}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{2}{11}\\\dfrac{1}{y+4}=\dfrac{-19}{11}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}2\left(x+1\right)=11x\\-19\left(y+4\right)=11\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x+2=11x\\-19y-76=11\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x-11x=-2\\-19y=87\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}-9x=-2\\y=\dfrac{-87}{19}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{2}{9}\left(tm\right)\\y=\dfrac{-87}{19}\left(tm\right)\end{matrix}\right.\)
- Thay x = \(\dfrac{2}{9}\) vào x+1 \(\ne\)0 có:
\(\dfrac{2}{9}+1\ne0\) \(\Leftrightarrow\dfrac{11}{9}\ne0\) (luôn đúng)
- Thay y = \(\dfrac{-87}{19}\) vào y+4 \(\ne\) 0có:
\(\dfrac{-87}{19}+4\ne0\) \(\Leftrightarrow\dfrac{-11}{19}\ne0\) (luôn đúng)
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