K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

20 tháng 9 2021

\(a,\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\ b,\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\23y=46\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ d,\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

\(e,\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

20 tháng 9 2021

a. \(\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=10\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x=20\\6x-2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y=46\\5x+2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4x+3y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

e. \(\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\4x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

28 tháng 9 2021

\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)

 

1 tháng 2 2023

\(a,\left\{{}\begin{matrix}2x-y=1\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}4x-2y=2\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}7x=7\\2x-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\2.1-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;1\right)\)

\(b,\left\{{}\begin{matrix}4x+3y=-1\\3x-2y=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}4.2x+3.2y=-1.2\\3.3x-2.3y=2.3\end{matrix}\right.\\ =>\left\{{}\begin{matrix}8x+6y=-2\\9x-6y=6\end{matrix}\right.\\ =>\left\{{}\begin{matrix}17x=4\\3x-2y=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=\dfrac{4}{17}\\y=-\dfrac{11}{17}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\dfrac{4}{17};-\dfrac{11}{17}\right)\)

13 tháng 1 2021

giúp mình nhé

30 tháng 3 2017

a) \(\left\{{}\begin{matrix}x+3y+2z=8\\2x+2y+z=6\\3x+y+z=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=2\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}x-3y+2z=-7\\-2x+4y+3z=8\\3x+y-z=5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{11}{14}\\y=\dfrac{5}{2}\\z=-\dfrac{1}{7}\end{matrix}\right.\)

5 tháng 5 2017

a) Đặt \(\left\{{}\begin{matrix}x+3y+2z=8\left(1\right)\\2x+2y+z=6\left(2\right)\\3x+y+z=6\left(3\right)\end{matrix}\right.\)
Cộng \(\left(2\right)+\left(3\right)\) ta có:\(\left\{{}\begin{matrix}x+3y+2z=8\left(1\right)\\2x+2y+z=6\left(2\right)\\5x+3y+2z=12\left(4\right)\end{matrix}\right.\)
Trừ \(\left(4\right)-\left(1\right)\) ta được: \(4x=4\Leftrightarrow x=1\).
Thay vào hệ phương trình ta được:
\(\left\{{}\begin{matrix}1+3y+2z=8\\2.1+2y+z=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\z=2\end{matrix}\right.\).
Vậy hệ phương trình có nghiệm: \(\left\{{}\begin{matrix}x=1\\y=1\\z=2\end{matrix}\right.\).

NV
28 tháng 1 2021

a.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

NV
28 tháng 1 2021

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)

22 tháng 9 2023

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)

a: Ta có: \(\left\{{}\begin{matrix}3x+2y=14\\5x+3y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15x+10y=70\\15x+9y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=67\\3x=14-2y=14-2\cdot67=-120\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\y=67\end{matrix}\right.\)

b: Ta có: \(\left\{{}\begin{matrix}-x+2y-6=0\\5x-3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x+10y=30\\5x-3y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7y=35\\2y-x=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)