Giải hệ phương trình \(\left\{{}\begin{matrix}\frac{y}{x}-\frac{y}{x+15}=\frac{1}{5}\\\frac{y}{x-3}-\frac{y}{x}=\frac{1}{20}\end{matrix}\right.\)
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giải hệ phương trình
1 , left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)+2xyleft(y-xright)left(y-1right)left(y+xright)left(y-2right)-2xyend{matrix}right.
2, left{{}begin{matrix}2left(frac{1}{x}+frac{1}{2y}right)+3left(frac{1}{x}-frac{1}{2y}right)^29left(frac{1}{x}+frac{1}{2y}right)-6left(frac{1}{x}-frac{1}{2y}right)^2-3end{matrix}right.
3 , left{{}begin{matrix}frac{xy}{x+y}frac{2}{3}frac{yz}{y+z}frac{6}{5}frac{zx}{z+x}frac{3}{4}end{matrix}right.
4 , left{{}begin...
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giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
Giải hệ phương trình :
1, left{{}begin{matrix}frac{2}{x}+frac{3}{y-2}4frac{4}{x}+frac{1}{y-2}1end{matrix}right.
2 , left{{}begin{matrix}frac{2}{2x-y}-frac{1}{x+y}0frac{3}{2x-y}-frac{6}{x+y}-1end{matrix}right.
3, left{{}begin{matrix}5left(x+2yright)3x-12x+43left(x-2yright)-15end{matrix}right.
4, left{{}begin{matrix}2x+y7-x+4y10end{matrix}right.
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Giải hệ phương trình :
1, \(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)
2 , \(\left\{{}\begin{matrix}\frac{2}{2x-y}-\frac{1}{x+y}=0\\\frac{3}{2x-y}-\frac{6}{x+y}=-1\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-2y\right)-15\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\)
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
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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...
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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\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
Giải các hệ phương trình:
\(a,\left\{{}\begin{matrix}\frac{1}{x}-\frac{1}{y}=1\\\frac{3}{x}+\frac{4}{y}=5\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{3}{4}\\\frac{1}{6x}+\frac{1}{5y}=\frac{2}{15}\end{matrix}\right.\)
a, ĐKXĐ : \(x,y\ne0\)
- Ta có : \(\left\{{}\begin{matrix}\frac{1}{x}-\frac{1}{y}=1\\\frac{3}{x}+\frac{4}{y}=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{3}{x}-\frac{3}{y}=3\\\frac{3}{x}+\frac{4}{y}=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{x}-\frac{1}{y}=1\\-\frac{7}{y}=-2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{x}-\frac{1}{\frac{2}{7}}=1\\y=\frac{2}{7}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{9}{7}\\y=\frac{2}{7}\end{matrix}\right.\)
Vậy phương trình có duy nhất 1 nghiệm là \(S=\left\{\frac{9}{7};\frac{2}{7}\right\}\)
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...
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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 , \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
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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...
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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.\(\left\{{}\begin{matrix}2x^2+5y^2=129\\-3x^2+y^2=13\end{matrix}\right.\)
Giải hệ phương trình
a, left{{}begin{matrix}sqrt[4]{x^3-1}+sqrt{x}3x^2+y^382end{matrix}right. d, left{{}begin{matrix}x-frac{1}{x}y-frac{1}{y}2yx^3+1end{matrix}right.
b, left{{}begin{matrix}sqrt{x+frac{1}{y}}+sqrt{x+y-3}32x+y+frac{1}{y}8end{matrix}right.
c,left{{}begin{matrix}frac{3}{x^2}2x+yfrac{3}{y^2}2y+xend{matrix}right.
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Giải hệ phương trình
a, \(\left\{{}\begin{matrix}\sqrt[4]{x^3-1}+\sqrt{x}=3\\x^2+y^3=82\end{matrix}\right.\) d, \(\left\{{}\begin{matrix}x-\frac{1}{x}=y-\frac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\sqrt{x+\frac{1}{y}}+\sqrt{x+y-3}=3\\2x+y+\frac{1}{y}=8\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}\frac{3}{x^2}=2x+y\\\frac{3}{y^2}=2y+x\end{matrix}\right.\)
Bài 2:
ĐK: ..........
Đặt $\sqrt{x+\frac{1}{y}}=a; \sqrt{x+y-3}=b$ $(a,b\geq 0$)
HPT \(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2+3=8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2=5\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ (a+b)^2-2ab=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ ab=2\end{matrix}\right.\)
Áp dụng định lý Vi-et đảo thì $a,b$ là nghiệm của pt $X^2-3X+2=0$
$\Rightarrow (a,b)=(2,1); (1,2)$
Nếu $(a,b)=(2,1)$
\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y-3=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y=4\end{matrix}\right.\Rightarrow y=\frac{1}{y}\Rightarrow y=\pm 1\)
$y=1\rightarrow x=3$
$y=-1\rightarrow y=5$
Nếu $(a,b)=(1,2)$
\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y-3=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y=7\end{matrix}\right.\Rightarrow y-\frac{1}{y}=6\)
\(\Rightarrow y^2-6y-1=0\Rightarrow y=3\pm \sqrt{10}\)
Nếu $y=3+\sqrt{10}\rightarrow x=4-\sqrt{10}$
Nếu $y=3-\sqrt{10}\rightarrow x=4+\sqrt{10}$
Vậy...........
Bài 1:
Đặt $\sqrt[4]{y^3-1}=a; \sqrt{x}=b$ $(a,b\geq 0$)
Khi đó hệ PT trở thành:
\(\left\{\begin{matrix} a+b=3\\ b^4+a^4+1=82\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^4+b^4=81\end{matrix}\right.\)
Có: \(a^4+b^4=81\)
\(\Leftrightarrow (a^2+b^2)^2-2a^2b^2=81\)
\(\Leftrightarrow [(a+b)^2-2ab]^2-2a^2b^2=81\)
\(\Leftrightarrow (9-2ab)^2-2a^2b^2=81\)
\(\Leftrightarrow 2a^2b^2-36ab=0\)
\(\Leftrightarrow ab(ab-18)=0\Rightarrow \left[\begin{matrix} ab=0\\ ab=18\end{matrix}\right.\)
Nếu $ab=0$. Kết hợp với $a+b=3$ suy ra $(a,b)=(3,0); (0,3)$
$\Rightarrow (x,y)=(0, \sqrt[4]{82}); (9, 1)$
Nếu $ab=18$. Kết hợp với $a+b=3$ và định lý Vi-et đảo suy ra $a,b$ là nghiệm của pt: $X^2-3X+18=0$
Dễ thấy pt này vô nghiệm nên loại
Vậy......
Bài 2:
ĐK: ..........
Đặt $\sqrt{x+\frac{1}{y}}=a; \sqrt{x+y-3}=b$ $(a,b\geq 0$)
HPT \(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2+3=8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2=5\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ (a+b)^2-2ab=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ ab=2\end{matrix}\right.\)
Áp dụng định lý Vi-et đảo thì $a,b$ là nghiệm của pt $X^2-3X+2=0$
$\Rightarrow (a,b)=(2,1); (1,2)$
Nếu $(a,b)=(2,1)$
\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y-3=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y=4\end{matrix}\right.\Rightarrow y=\frac{1}{y}\Rightarrow y=\pm 1\)
$y=1\rightarrow x=3$
$y=-1\rightarrow y=5$
Nếu $(a,b)=(1,2)$
\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y-3=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y=7\end{matrix}\right.\Rightarrow y-\frac{1}{y}=6\)
\(\Rightarrow y^2-6y-1=0\Rightarrow y=3\pm \sqrt{10}\)
Nếu $y=3+\sqrt{10}\rightarrow x=4-\sqrt{10}$
Nếu $y=3-\sqrt{10}\rightarrow x=4+\sqrt{10}$
Vậy...........
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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...
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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 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.
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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)\(\left\{{}\begin{matrix}x+y+xy=7\\x+y^2+xy=13\end{matrix}\right.\)