\(\left\{{}\begin{matrix}3x^2-13xy-10y^2=0\\2x^2-y^2+x=-22\end{matrix}\right.\)giải hệ pt sau
giải hệ pt :
a,\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3-30=0\\x^2y+x\left(1+y+y^2\right)+y-11=0\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}xy^2-2y+3x^2=0\\y^2+x^2y+2x=0\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}3xy+2y=5\\2xy\left(x+y\right)+y^2=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)
TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)
Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)
TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)
2 câu dưới hình như em hỏi rồi?
giải hệ pt:
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)
giúp mk vs ạ mai mk học rồi
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
giải hệ pt:
\(\left\{{}\begin{matrix}xy^2-2y+3x^2=0\\y^2+x^2y+2x=0\end{matrix}\right.\)
Với \(xy=0\) là nghiệm
Với \(xy\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}y-\dfrac{2}{x}+\dfrac{3x}{y}=0\\\dfrac{y}{x}+x+\dfrac{2}{y}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y-\dfrac{2}{x}=-\dfrac{3x}{y}\\x+\dfrac{2}{y}=-\dfrac{y}{x}\end{matrix}\right.\)
\(\Rightarrow\left(y-\dfrac{2}{x}\right)\left(x+\dfrac{2}{y}\right)=3\)
\(\Leftrightarrow xy-\dfrac{4}{xy}-3=0\)
\(\Rightarrow\left(xy\right)^2-3xy-4=0\Rightarrow\left[{}\begin{matrix}xy=-1\\xy=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{y}\\x=\dfrac{4}{y}\end{matrix}\right.\) thế vào \(y^2+x^2y+2x=0\)
\(\Rightarrow\left[{}\begin{matrix}y^2+\dfrac{1}{y}-\dfrac{2}{y}=0\\y^2+\dfrac{16}{y}+\dfrac{8}{y}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y^3=1\\y^3=-24\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải giúp mình vài hệ pt này nha
thanks nhiều
1.\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+x+\sqrt{y^2+x+y+1}+y=15\\\sqrt{x^2+x+y+1}-x+\sqrt{y^2+x+y+1}-y=2\end{matrix}\right.\)
2.\(\left\{{}\begin{matrix}\left(1-\dfrac{12}{y+3x}\right)\sqrt{x}=2\\\left(1+\dfrac{12}{y+3x}\right)\sqrt{y}=6\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}x^3+y^3=8\\x+y+2xy=2\end{matrix}\right.\)
4.\(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\)
5.\(\left\{{}\begin{matrix}x^3-3x=y^3-3y\\x^6+y^6=1\end{matrix}\right.\)
6.\(\left\{{}\begin{matrix}x^2-2xy+3y^2=9\\2x^2-13xy+15y^2=0\end{matrix}\right.\)
1. Đề này là 18 chứ không phải 15 nhé
\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+x+\sqrt{y^2+x+y+1}+y=18\left(1\right)\\\sqrt{x^2+x+y+1}-x+\sqrt{y^2+x+y+1}-y=2\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) và (1) - (2) ta được hệ mới
\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+\sqrt{y^2+x+y+1}=10\\x+y=8\end{matrix}\right.\)
\(\Rightarrow x=8-y\)
\(\Rightarrow\sqrt{x^2+9}+\sqrt{y^2+9}=10\)\(\Leftrightarrow\sqrt{x^2+9}=10-\sqrt{y^2+9}\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2+9=100-20\sqrt{y^2+9}+y^2+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\\left(8-y\right)^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\9y^2-72y+144=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
2. Dễ thấy x = y = 0 không phải là nghiệm của phương trình
HPT\(\Leftrightarrow\left\{{}\begin{matrix}1-\dfrac{12}{y+3x}=\dfrac{2}{\sqrt{x}}\left(1\right)\\1+\dfrac{12}{y+3x}=\dfrac{6}{\sqrt{y}}\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) ; (1) - (2) ta được
\(\left\{{}\begin{matrix}1=\dfrac{1}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}\left(3\right)\\\dfrac{12}{y+3x}=\dfrac{3}{\sqrt{y}}-\dfrac{1}{\sqrt{x}}\left(4\right)\end{matrix}\right.\)
Lấy ( 3) nhân (4)
\(\dfrac{12}{y+3x}=\dfrac{9}{y}-\dfrac{1}{x}=\dfrac{9x-y}{xy}\)
\(\Leftrightarrow27x^2-6xy-y^2=0\Leftrightarrow\left(9x+y\right)\left(3x-y\right)=0\)
\(\Rightarrow y=3x\)
đến đây thì dễ rồi
3. Đây là hệ đối xứng loại I
\(\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=8\\\left(x+y\right)+2xy=2\end{matrix}\right.\)
Đặt S = a + b ; P = ab (\(S^2\ge4P\) )
xong giải ra thôi mà
giải hệ pt sau
a\(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) b\(\left\{{}\begin{matrix}3x_{ }-2y=11\\4x-5y=3\end{matrix}\right.\) c\(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=_{ }-31\end{matrix}\right.\) D\(\left\{{}\begin{matrix}7X+5Y=19\\3x+5y=31\end{matrix}\right.\)
e\(\left\{{}\begin{matrix}7x-5y=3\\3x+10y=62\end{matrix}\right.\) f\(\left\{{}\begin{matrix}2x+5y=11\\3x+2y=11\end{matrix}\right.\) g\(\left\{{}\begin{matrix}x+3y=4y-x+5\\2x-y=3x-2\left(y+1\right)\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)
\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
giải hệ pt :
a, \(\left\{{}\begin{matrix}3y=\dfrac{y^2+2}{x^2}\\3x=\dfrac{x^2+2}{y^2}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^2y+xy^2+x-5y=0\\2xy+y^2-5y+1=0\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+y^2+xy+2y+x=2\\2x^2-y^2-2y-2=0\end{matrix}\right.\)
ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html
b.
Với \(xy=0\) không là nghiệm
Với \(xy\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y^2+1\right)=y\left(5-x^2\right)\\y^2+1=y\left(5-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y^2+1}{y}=\dfrac{5-x^2}{x}\\\dfrac{y^2+1}{y}=5-2x\end{matrix}\right.\)
\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)
\(\Leftrightarrow5-x^2=5x-2x^2\)
\(\Leftrightarrow...\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x\left(y+1\right)+\left(y+1\right)^2=3\\2x^2-\left(y+1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x\left(y+1\right)+\left(y+1\right)^2=3\\6x^2-3\left(y+1\right)^2=3\end{matrix}\right.\)
\(\Rightarrow5x^2-x\left(y+1\right)-4\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(5x+4\left(y+1\right)\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=-\dfrac{5x+4}{4}\end{matrix}\right.\)
Thế vào 1 trong 2 pt ban đầu...
Giải hệ phương trình sau bằng cách cộng hệ số
1) \(\left\{{}\begin{matrix}x-y=5\\2x+y=11\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}3x+2y=1\\3x+y=2\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}x-y=2\\3x+2y=11\end{matrix}\right.\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
giải hệ pt bằng phương pháp thế:
a,\(\left\{{}\begin{matrix}3x+y=-2\\-9x-39=6\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x+y=101\\-x+y=-1\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x+y=2\\\dfrac{1}{2}x+y=\dfrac{5}{4}\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x-5y=16\\10y-2x=-32\end{matrix}\right.\)
\(a.\left\{{}\begin{matrix}3x+y=-2\\-9x-39=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-2-3x\\-9x-36=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2-3x\\-9x=45\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-2-3x\\x=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=13\end{matrix}\right.\)
\(b.\left\{{}\begin{matrix}x+y=101\\-x+y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=101-y\\-x+y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=101-y\\-101+y+y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=101-y\\2y=100\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=101-y\\y=50\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=51\\y=50\end{matrix}\right.\)
\(c.\left\{{}\begin{matrix}x+y=2\\\dfrac{1}{2}x+y=\dfrac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2-y\\\dfrac{1}{2}x+y=\dfrac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2-y\\1-\dfrac{1}{2}y+y=\dfrac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2-y\\\dfrac{1}{2}y=\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2-y\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Giai hệ PT sau:\(\left\{{}\begin{matrix}2x^2+xy=3y+6\\2y^2+xy=3x+6\end{matrix}\right.\)
\(\left\{{}\begin{matrix}xy+x^2=1+y\\yx+y^2=1+x\end{matrix}\right.\)