\(\left\{{}\begin{matrix}\frac{4}{x-2y}-\frac{1}{x-2y}=1\\\frac{20}{x+2y}+\frac{3}{x-2y}=1\end{matrix}\right.\)
giup mh vs
a) \(\left\{{}\begin{matrix}2\left(x+y\right)+3(x-y)=4\\(x+y)+2(x-y)=5\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{6}{x-2y}+\frac{2}{x+2y}=3\\\frac{3}{x-2y}+\frac{4}{x+2y}=-1\end{matrix}\right.\)
a, Ta có : \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-\left(x-y\right)=4-10=-6\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=x-6\\\left(x+x-6\right)+2\left(x-x+6\right)=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=x-6\\x+x-6+12=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=x-6\\2x=-1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-\frac{1}{2}-6=-\frac{13}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là \(\left(x;y\right)=\left(-\frac{1}{2};-\frac{13}{2}\right)\)
b, ĐKXĐ : \(\left\{{}\begin{matrix}x-2y\ne0\\x+2y\ne0\end{matrix}\right.\)
=> \(x\ne\pm2y\)
- Ta có : \(\left\{{}\begin{matrix}\frac{6}{x-2y}+\frac{2}{x+2y}=3\\\frac{3}{x-2y}+\frac{4}{x+2y}=-1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{6}{x-2y}+\frac{2}{x+2y}=3\left(I\right)\\\frac{6}{x-2y}+\frac{8}{x+2y}=-2\left(II\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{12}{x-2y}+\frac{4}{x+2y}=6\left(III\right)\\\frac{3}{x-2y}+\frac{4}{x+2y}=-1\left(IV\right)\end{matrix}\right.\)
- Lấy ( I ) - ( II ) và ( III ) - ( IV ) ta được hệ phương trình :
\(\left\{{}\begin{matrix}-\frac{6}{x+2y}=5\\\frac{9}{x-2y}=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x+10y=-6\\7x-14y=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}35x+70y=-42\\35x-70y=45\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-\frac{6}{x+2y}=5\\140y=-87\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-\frac{6}{x-\frac{174}{140}}=5\\y=-\frac{87}{140}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x-\frac{870}{140}=-6\\y=-\frac{87}{140}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{3}{70}\\y=-\frac{87}{140}\end{matrix}\right.\)
Vậy hệ phương trình trên có nghiệm duy nhất là \(\left(x;y\right)=\left\{\frac{3}{70};-\frac{87}{140}\right\}\)
Giải hpt:
1, \(\left\{{}\begin{matrix}x^2+y+x^3y+x^2y+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^4+2x^2y+x^2y^2=-2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x-\frac{1}{x}=y-\frac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
3) ta xét phương trình thứ nhất
\(x-\frac{1}{x}=y-\frac{1}{y}\)
<=>\(x-y-\frac{1}{x}+\frac{1}{y}=0\)
<=>\(x-y-\left(\frac{1}{x}-\frac{1}{y}\right)=0\)
<=>\(x-y-\left(\frac{y-x}{xy}\right)=0\)
<=>\(\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\)
<=>\(x=y\) hoặc xy=-1
Với x=y thay vào phương trình thứ hai ta có
\(2x=x^3+1
\)
<=> \(x^3-2x+1=0\)
<=>\(x^3-x^2+x^2-x-x+1=0\)
<=>\(\left(x-1\right)\left(x^2+x-1\right)=0\)
<=> \(x=1\) hoặc \(x^2+x-1=0\)
\(x^2+x-1=0\) <=> \(x=\frac{-1+\sqrt{5}}{2}\)
hoặc \(x=\frac{-1-\sqrt{5}}{2}\)
Đối với xy=-1 thì y=-1/x thay vào phương trình 2 giải bình thường
Giải hpt sau:
a) \(\left\{{}\begin{matrix}\sqrt{5}x+\left(1-\sqrt{3}\right)y=1\\\left(1-\sqrt{3}\right)x+\sqrt{5}y=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{3x}{x+1}-\frac{2y}{y+4}=4\\\frac{2x}{x+1}-\frac{5y}{y+4}=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{2}{2x-y}+\frac{3}{x-2y}=\frac{1}{2}\\\frac{2}{2x-y}-\frac{1}{x-2y}=\frac{1}{18}\end{matrix}\right.\)
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 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{1-x+2y}{x-2y}+x+2y=4\\\frac{x+2y}{x-2y}-6=0\end{matrix}\right.\)
ĐKXĐ: \(x\ne2y\)
Biến đổi pt dưới:
\(x+2y-6\left(x-2y\right)=0\)
\(\Rightarrow5x=14y\Rightarrow x=\frac{14y}{5}\)
Thay vào pt trên:
\(\frac{1-\frac{14}{5}y+2y}{\frac{14}{5}y-2y}+\frac{14}{5}y+2y=4\)
\(\Leftrightarrow\frac{5}{4y}+\frac{24}{5}y-5=0\)
\(\Leftrightarrow\frac{24}{5}y^2-5y+\frac{5}{4}=0\) \(\Rightarrow\left[{}\begin{matrix}y=\frac{5}{12}\\y=\frac{5}{8}\end{matrix}\right.\) \(\Rightarrow x=\frac{14}{5}y=...\)
Giải hệ phương trình :
1, \(\left\{{}\begin{matrix}x-2y=1\\2x-y=4\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}-\frac{y}{y+12}=1\\\frac{x}{y+12}-\frac{x}{y}=2\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}3x^2+y^2=5\\x^2-3y=1\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\sqrt{3x-1}-\sqrt{2y+1}=1\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
a/ Bạn tự giải
b/ ĐKXĐ:...
Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)
Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)
\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)
\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)
Chắc bạn ghi sai đề, nghiệm quá xấu
3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)
4/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)
\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}2y\left(4y^2+3x^2\right)=x^4\left(x^2+3\right)\\2012^x\left(\sqrt{2y-2x+5}-x+1\right)=4024\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\\2x+\frac{1}{x+y}=1\end{matrix}\right.\)
\(2,\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\left(1\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(2y-x\right)\left(x^2-12y-15\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2y=x\\y=\frac{x^2-15}{12}\end{matrix}\right.\)
Ta xét các trường hợp sau:
Trường hợp 1:
\(y=\frac{x^2-15}{12}\) thay vào phương trình \(\left(2\right)\) ta được:
\(\frac{3x^2}{2\left(x^2-15\right)}+\frac{2x}{3}=\sqrt{\frac{4x^3}{x^2-15}+\frac{x^2}{4}}-\frac{x^2-15}{24}\)
\(\Leftrightarrow\frac{36x^2}{x^2-15}-12\sqrt{\frac{x^2}{x^2-15}\left(x^2+16x-15\right)}+\left(x^2+16x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\6\sqrt{\frac{x^2}{x^2-15}}=\sqrt{\left(x^2+16x-15\right)}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36\frac{x^2}{x^2-15}=x^2+16x-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\left(3\right)\end{matrix}\right.\)
Ta xét phương trình \(\left(3\right):36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\)
Vì: \(x=0\) Không phải là nghiệm. Ta chia cả hai vế p.trình cho \(x^2\) ta được:
\(36=\left(x-\frac{15}{x}\right)\left(x+16-\frac{15}{x}\right)\)
Đặt: \(x-\frac{15}{x}=t\Rightarrow t^2+16t-36=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-18\end{matrix}\right.\)
+ Nếu như:
\(t=2\Leftrightarrow x-\frac{15}{x}=2\Leftrightarrow x^2-2x-15=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)\(\Leftrightarrow x=5\)
+ Nếu như:
\(t=-18\Leftrightarrow x-\frac{15}{x}=-18\Leftrightarrow x^2+18x-15=0\Leftrightarrow\left[{}\begin{matrix}x=-9-4\sqrt{6}\\x=-9+4\sqrt{6}\end{matrix}\right.\Leftrightarrow x=-9-4\sqrt{6}\)
Trường hợp 2:
\(x=2y\) thay vào p.trình \(\left(2\right)\) ta được:
\(\Leftrightarrow\frac{x^2}{4x}+\frac{2x}{3}=\sqrt{\frac{2x^3}{3x}+\frac{x^2}{4}}-\frac{x}{4}\Leftrightarrow\frac{7}{6}x=\sqrt{\frac{11x^2}{12}}\Leftrightarrow x=0\left(ktmđk\right)\)
Vậy nghiệm của hệ đã cho là: \(\left(x,y\right)=\left(5;\frac{5}{6}\right),\left(-9-4\sqrt{6};\frac{27+12\sqrt{6}}{2}\right)\)
Năm mới chắc bị lag @@ tớ sửa luôn đề câu 3 nhé :v
3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\left(1\right)\\2xy+\frac{1}{x+y}=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow8\left[\left(x+y\right)^2-2xy\right]+4xy+\frac{5}{\left(x+y\right)^2}=13\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow8\left(a^2-2b\right)+4b+\frac{5}{a^2}=13\)
\(\Leftrightarrow8a^2-12b+\frac{5}{a^2}=13\)
Ta cũng có \(\left(2\right)\Leftrightarrow2b+\frac{1}{a}=1\)
\(\Leftrightarrow2b=1-\frac{1}{a}\)
Thay vào (1) ta được :
\(8a^2+\frac{5}{a^2}-6\cdot\left(1-\frac{1}{a}\right)=13\)
\(\Leftrightarrow8a^2+\frac{5}{a^2}-6+\frac{6}{a}=13\)
\(\Leftrightarrow8a^2+\frac{5}{a^2}+\frac{6}{a}=19\)
Giải pt được \(a=1\)
Khi đó \(b=\frac{1-\frac{1}{1}}{2}=0\)
Ta có hệ :
\(\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)
Vậy...
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
giải hệ phương trình
a)\(\left\{{}\begin{matrix}\frac{2}{x-1}+\frac{1}{2y+1}=\frac{6}{5}\\\frac{3}{x-1}-\frac{2}{2y+1}=\frac{11}{10}\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\\left(x+y\right)-3\sqrt{x+1}=-5\end{matrix}\right.\)
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 \(\left\{{}\begin{matrix}x+y=u\\\sqrt{x+1}=v\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u+v=4\\u-3v=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=1\\\sqrt{x+1}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x+1=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)