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a: \(\Leftrightarrow\left\{{}\begin{matrix}2x+2y+4z=8\\2x-y+3z=6\\2x-6y+8z=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y+z=2\\8y-4z=1\\x+y+2z=4\end{matrix}\right.\)
=>y=9/20; z=13/20; x=4-y-2z=9/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}z=23-x-y\\z=31-y-t\\z=27-t-x\\x+y+t=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-y+23=-y-t+31\\-y-t-31=-x-t+27\\x+y+t=33\\z=23-x-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x+t=8\\x-y=58\\x+y+t=33\\z=23-x-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=x+8\\y=x-58\\x-58+x+8+x=33\\z=23-x-y\end{matrix}\right.\)
=>x=83/3; t=107/3; y=-91/3; z=23-83/3+91/3=77/3
a: \(\Leftrightarrow\left\{{}\begin{matrix}4x+10y=6\\15x-10y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{34}{19}\\y=\dfrac{25}{19}\end{matrix}\right.\)
b: x+3y=5 và 2x-5y=-1
=>2x+6y=10 và 2x-5y=-1
=>11y=11 và x+3y=5
=>y=1 và x=2
c: 3x-4y=18 và 2x+y=1
=>3x-4y=18 và 8x+4y=4
=>11x=22 và 2x+y=1
=>x=2 và y=1-2*2=-3
a: Sửa đề:
\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5xz=6\left(z+x\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{4}{3}\\\dfrac{x+z}{xz}=\dfrac{5}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{4}{3}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{2}\\\dfrac{1}{y}=1\\\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{3};y=1;z=3\)
b: Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{7x-3y+2z}{7\cdot4-3\cdot3+2\cdot9}=\dfrac{37}{37}=1\)
=>x=4; y=3; z=9
Không mất tính tổng quát, giả sử \(x=min\left\{x;y;z\right\}\)
\(\Rightarrow z=3x^3+2x^2+x\le3y^3+2y^2+y\)
\(\Rightarrow z\le x\)
\(\Rightarrow z=x\)
\(\Rightarrow x=y=z\)
\(\Rightarrow x=3y^3+2x^2+x\Rightarrow x^2\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\end{matrix}\right.\)
1) \(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\8x+3\left(2-4x\right)=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=1\end{matrix}\right.\)
2) 2 pt 3 ẩn không giải được.
3) \(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\3x+2\left(x-2\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+1}{2}\\-4\cdot\frac{3y+1}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}2x+3y=5\\5x-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-3y+5}{2}\\5\cdot\frac{-3y+5}{2}-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}3x-y=7\\x+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2\left(3x-7\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}x+4y=2\\3x+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2-4y\\3\left(2-4y\right)+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{1}{5}\\x=\frac{6}{5}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}-x-y=2\\-2x-3y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-x-2\\-2x-3\left(-x-2\right)=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-15y+25x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-5\left(3y-5x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)
Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)
=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)
b: =>x^2-y^2-4y-2x-3=0 và x^2+2x+y=0
=>x^2-2x+1-y^2-4y-4=0 và x^2+2x+y=0
=>x=1 và y=-2 và x^2+2x+y=0
=>Hệ vô nghiệm
a: \(\Leftrightarrow\left\{{}\begin{matrix}z=2x-5\\y=3-2x+z=3-2x+2x-5=-2\\3x-2\cdot\left(-2\right)+2x-5=14\end{matrix}\right.\)
=>y=-2; 3x+4+2x-5=14; z=2x-5
=>y=-2; x=3; z=2*3-5=1
3: \(\left\{{}\begin{matrix}x+y+z=6\\2x-y+3z=9\\x+z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y+z+2x-y+3z=6+9\\x+z=4\\x+y+z=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+4z=15\\3x+3z=12\\x+y+z=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4z-3x-3z=15-12\\x+z=4\\y=6-x-z=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}z=3\\x=4-3=1\\y=2\end{matrix}\right.\)
5: \(\left\{{}\begin{matrix}2x+3y-z=11\\x-y+2z=-7\\x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y-z=11\\2x-2y+4z=-14\\2x-2y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+3y-z-2x+2y-4z=11+14\\2x+3y-z-2x+2y=11-6\\2x-2y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5y-5z=25\\5y-z=5\\x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y-5z-5y+z=25-5\\5y-z=5\\x-y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-4z=20\\5y=z+5\\x=y+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=-5\\y=\dfrac{z+5}{5}=\dfrac{-5+5}{5}=0\\x=0+3=3\end{matrix}\right.\)
8: \(\left\{{}\begin{matrix}2x+y=7\\x-y+2z=7\\z-3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+3y=21\\3x-3y+6z=21\\z-3y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+3y+3x-3y+6z=21+21\\6x+3y+z-3y=21-5\\z-3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+6z=42\\6x+z=16\\z-3y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+4z=28\\6x+z=16\\z-3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+4z-6x-z=28-16\\6x+z=16\\3y=z+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3z=12\\6x=16-z\\3y=z+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=4\\x=\dfrac{16-z}{6}=\dfrac{16-4}{6}=2\\y=\dfrac{z+5}{3}=\dfrac{4+5}{3}=3\end{matrix}\right.\)
\(3.\left\{{}\begin{matrix}x+y+z=6\\2x-y+3z=9\\x+z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6-4=2\\2x+3z=9+2=11\\x+z=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2\\2x+3z=11\\2x+2z=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\z=3\\x=4-3=1\end{matrix}\right.\\ 5.\left\{{}\begin{matrix}2x+3y-z=11\\x-y+2z=-7\\x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y-z=11\\2z=-7-3=-10\\x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=11-5=6\\z=\dfrac{-10}{2}=-5\\x-y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=6\\z=2\\2x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\z=2\\x-y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2\\z=2\\x=3+0=3\end{matrix}\right.\\ 8.\left\{{}\begin{matrix}2x+y=7\\x-y+2z=7\\z-3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\x-y+2z=7\\6y-2z=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\x+5y=17\\z-3y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+5y=35\\x+5y=17\\z-3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\x+5y=17\\z-3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\5y=17-2=15\\z=3y-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{15}{5}=3\\z=3\cdot3-5=4\end{matrix}\right.\)