Câu 1:
a, \(\left(\sqrt{18}+\frac{1}{2}\cdot12\sqrt{2}\right)\)/\(\sqrt{2}\)
b, \(\left\{{}\begin{matrix}3x-y=5\\x+y=-1\end{matrix}\right.\)
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) \(\left\{{}\begin{matrix}\frac{2}{2x-y}+\frac{3}{x-2y}=\frac{1}{2}\\\frac{2}{2x-y}-\frac{1}{x-2y}=\frac{1}{18}\end{matrix}\right.\)
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:
1, \(\left\{{}\begin{matrix}x^2+1+y^2+xy=y\\x+y-2=\frac{y}{1+x^2}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x^4+8y^4-2x-y=0\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x^2+y^2=\frac{1}{5}\\4x^2+3x-\frac{57}{25}=-y\left(3x+1\right)\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\sqrt{12-y}+\sqrt{y\left(12-x\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
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 pt:
a)\(\left\{{}\begin{matrix}x^2+y^2+x+y=18\\x\left(x+1\right).y\left(y+1\right)=72\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\3y-1=xy\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}2x+3y=xy+5\\\frac{1}{x}+\frac{1}{y+1}=1\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\sqrt{\frac{x}{y}}-3\sqrt{\frac{y}{x}}=2\\x-y+xy=1\end{matrix}\right.\) e)\(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
HELP ME :((
a/ \(\left\{{}\begin{matrix}\left(x^2+x\right)+\left(y^2+y\right)=18\\\left(x^2+x\right)\left(y^2+y\right)=72\end{matrix}\right.\)
Theo Viet đảo, \(x^2+x\) và \(y^2+y\) là nghiệm của:
\(t^2-18t+72=0\Rightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=12\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\left\{2;-3\right\}\\y=\left\{3;-4\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{3;-4\right\}\\y=\left\{2;-3\right\}\end{matrix}\right.\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\x=\frac{3y-1}{y}\end{matrix}\right.\)
Nhận thấy \(y=\frac{1}{3}\) không phải nghiệm
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\\frac{1}{x}=\frac{y}{3y-1}\end{matrix}\right.\) \(\Rightarrow\frac{y}{3y-1}+\frac{1}{y+1}=1\)
\(\Leftrightarrow y\left(y+1\right)+3y-1=\left(3y-1\right)\left(y+1\right)\)
\(\Leftrightarrow y^2-y=0\Rightarrow\left[{}\begin{matrix}y=0\left(l\right)\\y=1\end{matrix}\right.\) \(\Rightarrow x=2\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=xy+5\\y+1=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y-1=5\\y+1=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\y+1=xy\end{matrix}\right.\)
\(\Rightarrow y+1=\left(3-y\right)y\)
\(\Leftrightarrow y^2-2y+1=0\Rightarrow y=1\Rightarrow x=2\)
ai giúp t với
1:\(\left\{\begin{matrix}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
2:\(\left\{\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
3:\(\left\{\begin{matrix}y\left(x^2+2x+2\right)=x\left(y^2+6\right)\\\left(y-1\right)\left(x^2+2x+7\right)=\left(x+1\right)\left(y^2+1\right)\end{matrix}\right.\)
4:\(\left\{\begin{matrix}x-2\sqrt{y+1}=3\\x^3-4x^2\sqrt{y+1}-9x-8y=-52-4xy\end{matrix}\right.\)
5:\(\left\{\begin{matrix}\frac{y-2x+\sqrt{y}-x}{\sqrt{xy}}+1=0\\\sqrt{1-xy}+x^2-y^2=0\end{matrix}\right.\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\\sqrt{x}-\sqrt{y}=4,5\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\)
1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
4. Đk: \(x,y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)
giải các hpt sau: a)\(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}y=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{3x}{4}+\dfrac{2y}{5}=2,3\\x-\dfrac{3y}{5}=0,8\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)cíu zới
a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c: ĐKXĐ: y>2
\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)