\(\left\{{}\begin{matrix}\frac{x+1}{x-1}+\frac{3y}{y+2}=7\\\frac{2}{x-1}-\frac{5}{y+2}=4\end{matrix}\right.\)
Giải hệ phương trình trên bằng pp đặt ẩn phụ
Những câu hỏi liên quan
Giải hệ phương trình bằng phương pháp đặt ẩn phụ
\(\left\{{}\begin{matrix}\frac{5}{x+y-3}-\frac{2}{x-y+1}=8\\\frac{3}{x+y-2}+\frac{1}{x-y+1}=1,5\end{matrix}\right.\)
Cứ đặt ẩn phụ bình thường có sao đâu:
ĐKXĐ tự tìm
Đặt \(\left\{{}\begin{matrix}x+y-3=u\\x-y+1=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{5}{u}-\frac{2}{v}=8\\\frac{3}{u+1}+\frac{1}{v}=\frac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{u}-\frac{2}{v}=8\\\frac{6}{u+1}+\frac{2}{v}=3\end{matrix}\right.\)
\(\Rightarrow\frac{5}{u}+\frac{6}{u+1}=11\)
\(\Leftrightarrow5\left(u+1\right)+6u=11u\left(u+1\right)\)
\(\Leftrightarrow11u^2=5\Rightarrow u=\pm\sqrt{\frac{5}{11}}\Rightarrow v=...\)
Kết quả xấu quá, ko muốn làm tiếp nữa
Giải pt sau bằng cách đặt ẩn phụ
1, \(\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\frac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\frac{5}{4}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+3x^2-13x-15=\frac{8}{y^3}-\frac{8}{y}\\y^2+4=5y^2\left(x^2+2x+2\right)\end{matrix}\right.\)
a/ \(\left\{{}\begin{matrix}x^2+y+xy\left(x^2+y\right)+xy+1=-\frac{1}{4}\\x^4+y^2+2x^2y+xy+1=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y+1\right)\left(xy+1\right)=-\frac{1}{4}\\\left(x^2+y\right)^2+xy+1=-\frac{1}{4}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy+1=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\a^2+b=-\frac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\b=-\frac{1}{4}-a^2\end{matrix}\right.\)
\(\Rightarrow\left(a+1\right)\left(-\frac{1}{4}-a^2\right)=-\frac{1}{4}\)
\(\Leftrightarrow4a^3+4a^2+a=0\Leftrightarrow a\left(2a+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\Rightarrow b=-\frac{1}{4}\\a=-\frac{1}{2}\Rightarrow b=-\frac{1}{2}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x^2+y=0\\xy+1=-\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-x^2\\-x^3=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)
TH2: \(\left\{{}\begin{matrix}x^2+y=-\frac{1}{2}\\xy+1=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-\frac{1}{2}-x^2\\x\left(-\frac{1}{2}-x^2\right)=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)
Đúng 1
Bình luận (0)
b/ ĐKXĐ; ...
\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3x^2+3x+1-16x-16=\frac{8}{y^3}-\frac{8}{y}\\5\left(x^2+2x+2\right)=1+\frac{4}{y^2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\frac{8}{y^3}-\frac{8}{y}\\5\left(x+1\right)^2=\frac{4}{y^2}-4\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3-16a=8b^3-8b\\5a^2=4b^2-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^3-8b^3=16a-8b\\4=-5a^2+4b^2\end{matrix}\right.\)
Nhân vế với vế:
\(4\left(a^3-8b^3\right)=4\left(4a-2b\right)\left(-5a^2+4b^2\right)\)
\(\Leftrightarrow21a^3-10a^2b-16ab^2=0\)
\(\Leftrightarrow a\left(21a^2-10ab-16b^2\right)=0\)
\(\Leftrightarrow a\left(7a-8b\right)\left(3a+2b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\7a=8b\\3a=-2b\end{matrix}\right.\) \(\Rightarrow...\)
Đúng 1
Bình luận (0)
1.Giải hệ phương trình:
a.left{{}begin{matrix}2sqrt{2}x+y2sqrt{2}7x-3y7end{matrix}right.
b.left{{}begin{matrix}7x+y-frac{1}{7}-frac{4}{3}x-2y1frac{1}{3}end{matrix}right.
c.left{{}begin{matrix}2sqrt{5}x+3ysqrt{2}sqrt{5}x-y3sqrt{2}end{matrix}right.
d.left{{}begin{matrix}frac{2}{x}+frac{3}{y}-5frac{3}{x}-frac{4}{y}1end{matrix}right.
e.left{{}begin{matrix}-frac{5}{3x+1}+frac{7}{2x+1}frac{5}{7}frac{1}{3x+1}-frac{1}{2y-3}frac{2}{7}end{matrix}right.
g.left{{}begin{matrix}2x^2+5y^2129-3x^2+y^213en...
Đọc tiếp
1.Giải hệ phương trình:
a.\(\left\{{}\begin{matrix}2\sqrt{2}x+y=2\sqrt{2}\\7x-3y=7\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}7x+y=-\frac{1}{7}\\-\frac{4}{3}x-2y=1\frac{1}{3}\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}2\sqrt{5}x+3y=\sqrt{2}\\\sqrt{5}x-y=3\sqrt{2}\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y}=-5\\\frac{3}{x}-\frac{4}{y}=1\end{matrix}\right.\)
e.\(\left\{{}\begin{matrix}-\frac{5}{3x+1}+\frac{7}{2x+1}=\frac{5}{7}\\\frac{1}{3x+1}-\frac{1}{2y-3}=\frac{2}{7}\\\end{matrix}\right.\)
g.\(\left\{{}\begin{matrix}2x^2+5y^2=129\\-3x^2+y^2=13\end{matrix}\right.\)
hệ phương trình
1, left{{}begin{matrix}frac{1}{x+y}+frac{1}{x-y}frac{5}{8}frac{1}{x+y}-frac{1}{x-y}-frac{3}{8}end{matrix}right.
2, left{{}begin{matrix}frac{4}{2x-3y}+frac{5}{3x+y}2frac{3}{3x+y}-frac{5}{2x-3y}21end{matrix}right.
3, left{{}begin{matrix}frac{7}{x-y+2}+frac{5}{x+y-1}frac{9}{2}frac{3}{x-y+2}+frac{2}{x+y-1}4end{matrix}right.
4, left{{}begin{matrix}frac{3}{x}+frac{5}{y}-frac{3}{2}frac{5}{x}-frac{2}{y}frac{8}{3}end{matrix}right.
5 , left{{}begin{matrix}frac{2}{x+y-1}-frac{4}{x-y+1...
Đọc tiếp
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
hệ phương trình
1, left{{}begin{matrix}3x6x-3y2end{matrix}right.
2,left{{}begin{matrix}3x+5y152y-7end{matrix}right.
3, left{{}begin{matrix}7x-2y13x+y6end{matrix}right.
4, left{{}begin{matrix}3left(x+yright)+92left(x-yright)2left(x+yright)3left(x-yright)+11end{matrix}right.
5 , left{{}begin{matrix}3left(x+yright)+5left(x-yright)12-5left(x+yright)+2left(x-yright)11end{matrix}right.
6 , left{{}begin{matrix}2left(3x-2right)-45left(3y+2right)4left(3x-2right)+7left(3y+2right)-2end{matrix}right...
Đọc tiếp
hệ phương trình
1, \(\left\{{}\begin{matrix}3x=6\\x-3y=2\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}3x+5y=15\\2y=-7\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}7x-2y=1\\3x+y=6\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}3\left(x+y\right)+9=2\left(x-y\right)\\2\left(x+y\right)=3\left(x-y\right)+11\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}3\left(x+y\right)+5\left(x-y\right)=12\\-5\left(x+y\right)+2\left(x-y\right)=11\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}2\left(3x-2\right)-4=5\left(3y+2\right)\\4\left(3x-2\right)+7\left(3y+2\right)=-2\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{4}{5}\\\frac{1}{x}-\frac{1}{y}=\frac{1}{5}\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
có ái đó giúp mình với mình đang cần gấp
Đúng 0
Bình luận (0)
Giải các hệ phương trình sau
a)left{{}begin{matrix}frac{1}{x}+frac{1}{y+1}12x+3yxy+5end{matrix}right.
b)left{{}begin{matrix}left(x-yright)^2+3left(x-yright)42x+3y12end{matrix}right.
c)left{{}begin{matrix}frac{x}{y}+frac{y}{x}frac{13}{6}x+y5end{matrix}right.
d)left{{}begin{matrix}x+y+xy7x+y^2+xy13end{matrix}right.
Đọc tiếp
Giải các hệ phương trình sau
a)\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\2x+3y=xy+5\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(x-y\right)^2+3\left(x-y\right)=4\\2x+3y=12\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\frac{x}{y}+\frac{y}{x}=\frac{13}{6}\\x+y=5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}x+y+xy=7\\x+y^2+xy=13\end{matrix}\right.\)
hệ phương trình
1 ,left{{}begin{matrix}frac{2x-3}{2y-5}frac{3x+1}{3y-4}2left(x-3right)-3left(y+2right)-16end{matrix}right.
2, left{{}begin{matrix}frac{x}{y}frac{3}{2}3x-2y5end{matrix}right.
3, left{{}begin{matrix}frac{x^2-y-6}{x}x-2x+3y8end{matrix}right.
4, left{{}begin{matrix}frac{x}{y}frac{2}{3}x+y10end{matrix}right.
5, left{{}begin{matrix}frac{y^2+2x-8}{y}y-3x+y10end{matrix}right.
6 , left{{}begin{matrix}frac{x+1}{y-1}53left(2x-2right)-4left(3x+4right)5end{matrix}right.
7, left{{}begi...
Đọc tiếp
hệ phương trình
1 ,\(\left\{{}\begin{matrix}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{3}{2}\\3x-2y=5\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{x^2-y-6}{x}=x-2\\x+3y=8\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{2}{3}\\x+y=10\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\frac{y^2+2x-8}{y}=y-3\\x+y=10\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}\frac{x+1}{y-1}=5\\3\left(2x-2\right)-4\left(3x+4\right)=5\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}2x+y=4\\\left|x-2y\right|=3\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{2x}{x+1}+\frac{y}{y+1}=3\\\frac{x}{x+1}-\frac{3y}{y+1}=-1\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}y-\left|x\right|=1\\2x-y=1\end{matrix}\right.\)
10 , \(\left\{{}\begin{matrix}\sqrt{x+3y}=\sqrt{3x-1}\\5x-y=9\end{matrix}\right.\)
Giải các hệ phương trình
1. \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{2}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{4}{y^2}=7\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}2x+3y=x^2+3xy+y^2\\x^2+2y^2=x+2y\end{matrix}\right.\)
@Nguyễn Việt Lâm
câu 1 thôi 2 làm được rồi
Ta có: \(\left(2x+3y\right)\left(x^2+2y^2\right)=\left(x^2+3xy+y^2\right)\left(x+2y\right)\)
\(\Leftrightarrow2x^3+4xy^2+3x^2y+6y^3=x^3+2x^2y+3x^2y+6xy^2+x^2y+2y^3\)
\(\Leftrightarrow x^3-2xy^2-3x^2y+4y^3=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2-4xy-4y^2\right)=0\)
Xem thêm câu trả lời
Giải hệ phương trình bằng phương pháp đặt ẩn phụ
\(\left\{{}\begin{matrix}\frac{2x}{1-3x}+\frac{y}{y+1}=7\\\frac{5x}{1-3x}-\frac{4y}{y+1}=-2\end{matrix}\right.\)
