Giải hệ phương trình:
\(\left\{{}\begin{matrix}x^2+xy=\frac{16}{x}\\xy+y^2=\frac{16}{y}\end{matrix}\right.\)
giải phương trình \(\left\{{}\begin{matrix}xy-\frac{x}{y}\\xy-\frac{y}{x}=\frac{9}{2}\end{matrix}\right.=\frac{16}{3}}\)
Giải hệ phương trình sau
a. \(\left\{{}\begin{matrix}x^2+y^2+2\left(x+y\right)=23\\x+y+xy=11\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}x^2+4x+y=0\\\left(x+2\right)^2+5y=16\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=9\end{matrix}\right.\)
Giải hệ pt:
\(\left\{{}\begin{matrix}\frac{1}{2}xy+18=\frac{1}{2}\left(x+2\right)\left(y+2\right)\\\frac{1}{2}xy-16=\frac{1}{2}\left(x-2\right)\left(y+3\right)\end{matrix}\right.\)
Hệ phương trình đề cho tương đương
\(\left\{{}\begin{matrix}\frac{1}{2}xy+18=\frac{1}{2}xy+x+y+2\\\frac{1}{2}xy-16=\frac{1}{2}xy+\frac{3}{2}x-y-3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2=18\\\frac{3}{2}x-y-3=-16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=16\\\frac{3}{2}x-y=-13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{3}{2}x=3\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{6}{5}\\y=\frac{74}{5}\end{matrix}\right.\)
KL: ........................
Giải hệ phương trình
a)\(\left\{{}\begin{matrix}2x^3=y+1\\2y^3=x+1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=7\\x^2-\frac{1}{y^2}=3\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+y^2=10\\x+y=4\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}xy+x+y=19\\x^2y+xy^2=84\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2+xy+y^2=4\\x+xy+y=2\end{matrix}\right.\)
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)\(\left\{{}\begin{matrix}x+y+xy=7\\x+y^2+xy=13\end{matrix}\right.\)
giải các hệ phương trình sau:
1, \(\left\{{}\begin{matrix}\left(x+2\right)\left(y-2\right)=xy\\\left(x+4\right)\left(y-3\right)=xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{1}{x-3}+\frac{1}{y}=2\\\frac{2}{x-3}-\frac{3}{y}=1\end{matrix}\right.\)
1, \(\left\{{}\begin{matrix}\left(x+2\right)\left(y-2\right)=xy\left(1\right)\\\left(x+4\right)\left(y-3\right)=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-2x+2y-4=xy\\xy-3x+4y-12=xy\end{matrix}\right.\)
\(\Rightarrow x-2y+8=0\Leftrightarrow x=2y-8\) thay vào \(\left(1\right)\) ta được
\(\left(2y-6\right)\left(y-2\right)=\left(2y-8\right)y\)\(\Leftrightarrow2y^2-4y-6y+12=2y^2-8y\Leftrightarrow2y=12\Leftrightarrow y=6\Rightarrow x=4\)
Vậy hệ phương trình có nghiệm là \(\left(x,y\right)=\left(4,6\right)\)
Câu 1 nhân 2 tích đó vào, rồi ra tích x.y xong rút gọn x.y ra lại hệ pt quen thuộc.
Câu 2 đặt ẩn, \(\frac{1}{x-3}=a\) và \(\frac{1}{y}=b\)
lại ra hpt quen thuộc, giải a ,b xong thay vào tìm x với y
giải hệ: a, \(\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\sqrt[]{x-1}+\sqrt[]{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x\sqrt[]{x}+y\sqrt[]{y}=35\\x\sqrt[]{y}+y\sqrt[]{x}=30\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x^2+xy+y^2=3\\x+xy+y=-1\end{matrix}\right.\)
e,\(\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\)
\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)
Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)
Làm nốt nha
\(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\left(x;y\ge1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=4\\x+y=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-\left(x+y\right)+1}=6\\x+y=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-xy+1}=6\\x+y=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=4\end{matrix}\right.\)
Làm nốt
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
\(\left\{{}\begin{matrix}\frac{x+y}{xy}+\frac{xy}{x+y}=\frac{5}{2}\\\frac{x-y}{xy}+\frac{xy}{x-y}=\frac{10}{3}\end{matrix}\right.\)
Đặt\(\frac{x+y}{xy}\)=a =>\(\frac{xy}{x+y}\)=\(\frac{1}{a}\)
và \(\frac{x-y}{xy}\)=b =>\(\frac{xy}{x-y}=\frac{1}{b}\)
Ta có:\(\left\{{}\begin{matrix}a+\frac{1}{a}=\frac{5}{2}\left(1\right)\\b+\frac{1}{b}=\frac{10}{3}\left(2\right)\end{matrix}\right.\)
(1)<=>\(2a^2-5a+2=0\)
<=>\(\left(2a-1\right)\left(a-2\right)=0\)
=>\(a=2\) hoặc \(a=\frac{1}{2}\)
=>\(xy=\frac{x+y}{2}\) hoặc \(xy=2\left(x+y\right)\)(3)
Tương tự (2)có:\(\left(3b-1\right)\left(b-3\right)=0\)
<=>\(b=\frac{1}{3}\) hoặc \(b=3\)
=>\(xy=\frac{x-y}{3}\) hoặc \(xy=3\left(x-y\right)\)(4)
Từ (3) và (4) tự tính nghiệm nha