\(\left\{{}\begin{matrix}x+2y=10\\\dfrac{1}{2}x-y=1\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=2\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
Những câu hỏi liên quan
giả các hệ phương trình sau :a) left{{}begin{matrix}dfrac{-3}{x-y+1}+dfrac{1}{x
+y-2}12dfrac{2}{x-y+1}-dfrac{3}{x+y-2}-1end{matrix}right.b) left{{}begin{matrix}x^2+2left(y^2+2yright)103x^2-left(y^2+2yright)9end{matrix}right.c) left{{}begin{matrix}dfrac{7}{sqrt{x-1}}-dfrac{5}{sqrt{y+2}}dfrac{9}{2}dfrac{3}{sqrt{x-1}}+dfrac{2}{sqrt{y+2}}4end{matrix}right.
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giả các hệ phương trình sau :
a) \(\left\{{}\begin{matrix}\dfrac{-3}{x-y+1}+\dfrac{1}{x +y-2}=12\\\dfrac{2}{x-y+1}-\dfrac{3}{x+y-2}=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+2\left(y^2+2y\right)=10\\3x^2-\left(y^2+2y\right)=9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x-1}}-\dfrac{5}{\sqrt{y+2}}=\dfrac{9}{2}\\\dfrac{3}{\sqrt{x-1}}+\dfrac{2}{\sqrt{y+2}}=4\end{matrix}\right.\)
1)left{{}begin{matrix}2x+dfrac{1}{y}dfrac{3}{x}2y+dfrac{1}{x}dfrac{3}{y}end{matrix}right.2)left{{}begin{matrix}x^33x+8yy^33y+8xend{matrix}right.3)left{{}begin{matrix}x^2+y^2+x-2y2x^2+y^2+2x+2y11end{matrix}right.4)left{{}begin{matrix}x^3-y13x^2-3xy+y^21end{matrix}right.5)left{{}begin{matrix}x^3-y^39left(x-yright)left(x^2+y^2right)15end{matrix}right.
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1)\(\left\{{}\begin{matrix}2x+\dfrac{1}{y}=\dfrac{3}{x}\\2y+\dfrac{1}{x}=\dfrac{3}{y}\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}x^3=3x+8y\\y^3=3y+8x\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x^2+y^2+x-2y=2\\x^2+y^2+2x+2y=11\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^3-y=1\\3x^2-3xy+y^2=1\end{matrix}\right.\)
5)\(\left\{{}\begin{matrix}x^3-y^3=9\\\left(x-y\right)\left(x^2+y^2\right)=15\end{matrix}\right.\)
Giải hệ phương trình:
a)left{{}begin{matrix}dfrac{3x+2}{x-1}-dfrac{3y-1}{y+2}0dfrac{2}{x-1}+dfrac{3}{y+2}1end{matrix}right.
b)left{{}begin{matrix}dfrac{4x-5}{x+1}+dfrac{2y-3}{y-5}8dfrac{3}{x+1}-dfrac{2}{y-5}-1end{matrix}right.
c)left{{}begin{matrix}dfrac{x+y-2}{x+1}+dfrac{3-x}{y+1}dfrac{5}{4}dfrac{3left(x+y-2right)}{x+1}-dfrac{5-x+2y}{y+1}dfrac{3}{4}end{matrix}right.
d)left{{}begin{matrix}dfrac{x-y+1}{x-3}+dfrac{x+1}{y-3}dfrac{-7}{2}dfrac{2left(x-y+1right)}{x-3}-dfrac{x+y-2}{y-3}-dfrac{9}{2}...
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Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
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}1end{matrix}right.
2) left{{}begin{matrix}dfrac{2}{x+2y}+dfrac{1}{y+2x}3dfrac{4}{x+2y}-dfrac{3}{y+2x}1end{matrix}right.
3) left{{}begin{matrix}dfrac{3x}{x+1}-dfrac{2}{y+4}4dfrac{2x}{x+1}-dfrac{5}{y+4}9end{matrix}right.
4) left{{}begin{matrix}x^2+y^2133x^2-2y^2-6end{matrix}right.
5) left{{}begin{matrix}3sqrt{x}+2sqrt{y}162sqrt{x}-3sqrt{y}-11end{matrix}right....
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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?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
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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)
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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 :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\) \(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}-7b=\dfrac{-1}{3}\\a+b=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a+\dfrac{1}{21}=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\left(tm\right)\\a=\dfrac{1}{28}\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{28}\\\dfrac{1}{y}=\dfrac{1}{21}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
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Xem thêm câu trả lời
giải hệ pt:9) left{{}begin{matrix}dfrac{7}{2x+y}+dfrac{4}{2x-y}74dfrac{3}{2x+y}+dfrac{2}{2x-y}32end{matrix}right.10) left{{}begin{matrix}x2y-12x-y5end{matrix}right.11) left{{}begin{matrix}3x-602y-x4end{matrix}right.12) left{{}begin{matrix}2x+y5x+7y9end{matrix}right.13) left{{}begin{matrix}dfrac{3}{x}-dfrac{4}{y}2dfrac{4}{x}-dfrac{5}{y}3end{matrix}right.14) left{{}begin{matrix}dfrac{1}{x}+dfrac{1}{y}dfrac{1}{12}dfrac{8}{x}+dfrac{15}{y}1end{matrix}right.15) left{{}begin{matrix}2sqrt{x-1}-sqrt{y-1}...
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giải hệ pt:
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)
giúp mk vs ạ mai mk học rồi
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
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13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
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3 tháng 12 2021 lúc 20:17
Giải hệ phương trìnha)left{{}begin{matrix}x+ydfrac{x-3}{2}x+2ydfrac{2-4y}{15}end{matrix}right. b)left{{}begin{matrix}dfrac{1}{x}+dfrac{1}{y}-1dfrac{3}{x}-dfrac{2}{y}7end{matrix}right.c)left{{}begin{matrix}sqrt{x+3}-2sqrt{y+1}22sqrt{x+3}+sqrt{y+1}4end{matrix}right. d)left{{}begin{matrix}dfrac{7}{sqrt{x}-7}-dfrac{4}{sqrt{y}+6}dfrac{5}{3}dfrac{5}{sqrt{x}-7}+dfrac{3}{sqrt{y}+6}2dfrac{1}{9}end{matrix}right.
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Giải hệ phương trình
a)\(\left\{{}\begin{matrix}x+y=\dfrac{x-3}{2}\\x+2y=\dfrac{2-4y}{15}\end{matrix}\right.\) b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\) d)\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{9}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
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giải hệ phương trình
a,left{{}begin{matrix}dfrac{2}{x}+dfrac{3}{y}5dfrac{1}{x}-dfrac{4}{y}-3end{matrix}right.
b,left{{}begin{matrix}dfrac{12}{x-3}-dfrac{5}{y+2}63dfrac{8}{x-3}+dfrac{15}{y+2}-13end{matrix}right.
c,left{{}begin{matrix}dfrac{4}{x+2}-dfrac{1}{x-2y}1dfrac{20}{x+2y}+dfrac{3}{x-2y}1end{matrix}right.
d,left{{}begin{matrix}left|x-1right|+left|y-2right|2left|x-1right|+y3end{matrix}right.
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giải hệ phương trình
a,\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}\dfrac{4}{x+2}-\dfrac{1}{x-2y}=1\\\dfrac{20}{x+2y}+\dfrac{3}{x-2y}=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-2\right|=2\\\left|x-1\right|+y=3\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
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Giải hệ
a) \(\left\{{}\begin{matrix}x^2+y^2-2y-6+2\sqrt{2y+3}=0\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2y+2y+x=4xy\\\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{x}{y}=3\end{matrix}\right.\)
10. giải hpt bằng phương pháp thế:6) left{{}begin{matrix}2y-403x+y-4end{matrix}right.7) left{{}begin{matrix}4x-6y2x-dfrac{3}{2}ydfrac{1}{2}end{matrix}right.8) left{{}begin{matrix}dfrac{x}{3}+dfrac{y}{2}12x+3ydfrac{2}{5}end{matrix}right.9) left{{}begin{matrix}3x-2y2x+3y6end{matrix}right.10) left{{}begin{matrix}2x+3y24x-y-10end{matrix}right.11) left{{}begin{matrix}3x-2y32x-dfrac{4}{3}y1end{matrix}right.12) left{{}begin{matrix}5x+y32x+0,4y1,2end{matrix}right.giúp mk vs ạ mai mk học rồi
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10. giải hpt bằng phương pháp thế:
6) \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}3x-2=y\\2x+3y=6\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}2x+3y=2\\4x-y-1=0\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-2y=3\\2x-\dfrac{4}{3}y=1\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}5x+y=3\\2x+0,4y=1,2\end{matrix}\right.\)
giúp mk vs ạ mai mk học rồi
6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\\dfrac{2+6y}{4}-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-2\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y
(Các câu khác tương tự nhé.)
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