Giải HPT: \(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=2\\\left(x-3\right)\left(y+1\right)=-6\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-1\right).6+\left(y-2\right).\left(-2\right)=0\\\left(x+1\right).4+\left(y-1\right).\left(-3\right)=0\end{matrix}\right.\)
Giải hpt
\(\Leftrightarrow\left\{{}\begin{matrix}6x-6-2y+4=0\\4x+4-3y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=1\\4x-3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
Giải các hpt:
1)\(\left\{{}\begin{matrix}\frac{10}{x-1}+\frac{1}{y+2}=1\\\frac{25}{x-1}+\frac{3}{y+2}=2\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}4\left|x+y\right|-3\left|x-y\right|=8\\3\left|x+y\right|-5\left|x-y\right|=6\end{matrix}\right.\)
1. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}x-y=4\\3x+4y=19\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}x-\sqrt{3y}=\sqrt{3}\\\sqrt{3x}+y=7\end{matrix}\right.\)
2. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}2-\left(x-y\right)-3\left(x+y\right)=5\\3\left(x-y\right)+5\left(x+y\right)=-2\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x+y=24\\\dfrac{x}{9}+\dfrac{y}{27}=2\dfrac{8}{9}\end{matrix}\right.\) d, \(\left\{{}\begin{matrix}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2=15}\end{matrix}\right.\)
3. Cho hpt \(\left\{{}\begin{matrix}\left(m+1\right)x-y=3\\mx+y=m\end{matrix}\right.\)
a, Giải hpt khi m=\(\sqrt{2}\)
b, tìm giá trị của m để hpt có nghiệm duy nhất thỏa mãn: x+y>0
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
giải hpt: a,\(\left\{{}\begin{matrix}x+y-\sqrt{xy}=3\\\sqrt{x+1}+\sqrt{y+1}=4\end{matrix}\right.\) b,\(\left\{{}\begin{matrix}x+y=5+\sqrt{\left(x-1\right)\left(y-1\right)}\\\sqrt{x-1}+\sqrt{y-1}=3\end{matrix}\right.\)
a.
ĐKXĐ: \(x;y\ge-1;xy\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-3=\sqrt{xy}\\x+y+2\sqrt{xy+x+y+1}=14\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\ge0\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}u-3=\sqrt{v}\\u+2\sqrt{u+v+1}=14\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-6u+9\left(u\ge3\right)\\4\left(u+v+1\right)=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\4u+4\left(u^2-6u+9\right)+4=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\3u^2+8u-156=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\\left[{}\begin{matrix}u=6\\u=-\dfrac{26}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=6\\v=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=6\\xy=9\end{matrix}\right.\) \(\Rightarrow x=y=3\)
b.
ĐKXĐ: \(x;y\ge1\)
Xét \(\sqrt{x-1}+\sqrt{y-1}=3\)
\(\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=\dfrac{11-x-y}{2}\)
Thế vào pt đầu:
\(x+y=5+\dfrac{11-x-y}{2}\)
\(\Leftrightarrow x+y=7\Rightarrow y=7-x\)
Thế xuống pt dưới:
\(\sqrt{x-1}+\sqrt{6-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x-1\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\left(x-1\right)\left(6-x\right)=4\)
\(\Leftrightarrow...\)
giải hpt
\(\left\{{}\begin{matrix}2\left(x+3\right)-5\left(y-1\right)=3\\5\left(x+3\right)+\left(y-1\right)=6\end{matrix}\right.\)
giải hpt \(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-3\right)^2=1\\\left(x-1\right)\left(y-3\right)+3=x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-3\right)^2=1\\\left(x-1\right)\left(y-3\right)-\left(x-1\right)-\left(y-3\right)+1=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x-1=a\\y-3=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=1\\ab-a-b+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\left(a-1\right)\left(b-1\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\\\left\{{}\begin{matrix}a=0\\b=1\end{matrix}\right.\end{matrix}\right.\)
Giải HPT :
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x+1\right)\left(y+1\right)=8\\7y^3+6xy\left(x+2y\right)=25\end{matrix}\right.\)
giải hpt: \(\left\{{}\begin{matrix}\left(2x-1\right)^2+4\left(y-1\right)^2=10\\xy\left(x-1\right)\left(y-2\right)=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x^2-x\right)+1+4\left(y^2-2y\right)+4=10\\\left(x^2-x\right)\left(y^2-2y\right)=-\dfrac{3}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-x=u\\y^2-2y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4u+1+4v+4=10\\uv=-\dfrac{3}{2}\end{matrix}\right.\)
Chắc em tự giải được hệ này, chỉ cần thế là xong
giải hpt \(\left\{{}\begin{matrix}3x-2\left|y\right|=1\\x+3\left|y\right|=4\end{matrix}\right.\)
=>3x-2|y|=1 và 3x+9|y|=12
=>-11|y|=-11 và x+3|y|=4
=>x=1 và |y|=1
=>x=1 và \(y\in\left\{1;-1\right\}\)