Cho \(\left\{{}\begin{matrix}\dfrac{1}{3}< x\le\dfrac{1}{2}\\y\ge1\end{matrix}\right.\). Tìm Min \(P=x^2+y^2+\dfrac{x^2y^2}{\left(\left(4x-1\right)y-x\right)^2}\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
1)\(\left\{{}\begin{matrix}2x+\dfrac{1}{y}=\dfrac{3}{x}\\2y+\dfrac{1}{x}=\dfrac{3}{y}\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}x^3=3x+8y\\y^3=3y+8x\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x^2+y^2+x-2y=2\\x^2+y^2+2x+2y=11\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^3-y=1\\3x^2-3xy+y^2=1\end{matrix}\right.\)
5)\(\left\{{}\begin{matrix}x^3-y^3=9\\\left(x-y\right)\left(x^2+y^2\right)=15\end{matrix}\right.\)
giải hệ phương trình
a,\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}\dfrac{4}{x+2}-\dfrac{1}{x-2y}=1\\\dfrac{20}{x+2y}+\dfrac{3}{x-2y}=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-2\right|=2\\\left|x-1\right|+y=3\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
Giải hpt:
a)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}\dfrac{-5x+2y}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(-5x+2y\right)+60=3\left(y+27\right)-24x\\7\left(x+1\right)+21y=3\left(6y-5x\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-20x+8y+60=3y+81-24x\\7x+7+21y=18y-15x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-20x+8y-3y+24x=21\\7x+21y-18y+15x=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+5y=21\\22x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+15y=63\\110x+15y=-35\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-98x=98\\4x+5y=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\5y=21-4x=21+4=25\end{matrix}\right.\)
=>x=-1 và y=5
b: \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(xy+3x+2y+6\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(xy-2x-2y+4\right)=32\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}xy+3x+2y+6-xy=100\\xy-\left(xy-2x-2y+4\right)=64\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+2y=94\\2x+2y=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=34\\2x+2y=60\end{matrix}\right.\)
=>x=34 và y=-4
c: \(\left\{{}\begin{matrix}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{matrix}\right.\)
\(\left\{{}\begin{matrix}xy-x+20y-20=xy\\xy+x-10y-10=xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x+20y=20\\x-10y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10y=30\\x-10y=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x=10y+10=30+10=40\end{matrix}\right.\)
d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-2y\\x< >-\dfrac{y}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{2x+y}=3\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x+2y}+\dfrac{2}{2x+y}=6\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{2x+y}=5\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+y=1\\\dfrac{4}{x+2y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=1\\x+2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+y=1\\2x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=1\\x+2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=1-2y=1-\dfrac{2}{3}=\dfrac{1}{3}\end{matrix}\right.\)(nhận)
e: ĐKXĐ: x<>-1 và y<>-4
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=\dfrac{27}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\left(nhận\right)\)
Giải hệ
a) \(\left\{{}\begin{matrix}x^2+y^2-2y-6+2\sqrt{2y+3}=0\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2y+2y+x=4xy\\\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{x}{y}=3\end{matrix}\right.\)
giải giúp mik bt này vs mn!
1)\(\left\{{}\begin{matrix}2x^2+y^2+x=3\left(xy+1\right)+2y\\\dfrac{2}{3+\sqrt{2x-y}}+\dfrac{2}{3+\sqrt{4-5x}}=\dfrac{9}{2x-y+9}\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}\left(x+3y+1\right)\sqrt{2xy+2y}=y\left(3x+4y+3\right)\\\left(\sqrt{x+3}-\sqrt{2y-2}\right)\left(x-3+\sqrt{x^2+x+2y-4}\right)=4\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}\sqrt{2x-3}=\left(y^2+2011\right)\left(5-y\right)+\sqrt{y}\\y\left(y-x+2\right)=3x+3\end{matrix}\right.\)
5)\(\left\{{}\begin{matrix}x^3+2x^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14=x-2}\end{matrix}\right.\)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
giả các hệ phương trình sau :
a) \(\left\{{}\begin{matrix}\dfrac{-3}{x-y+1}+\dfrac{1}{x +y-2}=12\\\dfrac{2}{x-y+1}-\dfrac{3}{x+y-2}=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+2\left(y^2+2y\right)=10\\3x^2-\left(y^2+2y\right)=9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x-1}}-\dfrac{5}{\sqrt{y+2}}=\dfrac{9}{2}\\\dfrac{3}{\sqrt{x-1}}+\dfrac{2}{\sqrt{y+2}}=4\end{matrix}\right.\)
giải các hệ phương tình sau :
1) \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=9\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x\left(3x+2y\right)\left(x+1\right)=12\\x^2+2y+4x-8=0\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}x-3y=\dfrac{4y}{x}\\y-3x=\dfrac{4x}{y}\end{matrix}\right.\)
giúp mình với ạ ><
1. \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=9\end{matrix}\right.\) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y+xy^2+x+y=5xy\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4y^2+x^2y^4+x^2+y^2=25x^2y^2\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\)\(\Leftrightarrow0=16x^2y^2\)
\(\Rightarrow\) phương trình vô nghiệm