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Mấy bài này đơn giản , bạn chỉ cần rút x hoặc y ra là đc
mk làm ví dụ một câu ha
\(\left\{{}\begin{matrix}x+2y=1\\-3x-y=2\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=1-2y\left(1\right)\\-3x-y=2\left(2\right)\end{matrix}\right.\)
Thay (1) vào bt (2) ta có -3(1-2y)-y=2
Bạn giải ra y rồi giải ra x là xong
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)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
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b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
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c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)
\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)
Vậy hệ pt vô nghiệm
d) Nhân 3 pt đầu rồi thu gọn
a) \(\left\{{}\begin{matrix}2x-y=5\left(1\right)\\x+y=4\left(2\right)\end{matrix}\right.\)
Lấy (1)+(2) theo vế ta được \(\left(2x-y\right)+\left(x+y\right)=5+4\Leftrightarrow3x=9\Leftrightarrow x=3\)
Thay x=3 vào (2) được \(3+y=4\Leftrightarrow y=1\)
Vậy (x;y)=(3;1)
b) \(\left\{{}\begin{matrix}x+y=7\left(3\right)\\x-y=3\left(4\right)\end{matrix}\right.\)
Lấy (3)+(4) theo vế ta được \(\left(x+y\right)+\left(x-y\right)=7+3\Leftrightarrow2x=10\Leftrightarrow x=5\)
Thay x=5 vào (3) được \(5+y=7\Leftrightarrow y=2\)
Vậy (x;y)=(5;2)
c) \(\left\{{}\begin{matrix}x-2y=4\left(5\right)\\x+y=1\left(6\right)\end{matrix}\right.\)
Lấy (6)-(5) theo vế ta được \(\left(x+y\right)-\left(x-2y\right)=1-4\Leftrightarrow3y=-3\Leftrightarrow y=-1\)
Thay y=-1 vào (6) được \(x+\left(-1\right)=1\Leftrightarrow x=2\)
Vậy (x;y)=(2;-1)
a)\(\left\{{}\begin{matrix}x=7-2y\\3\left(7-2y\right)-4y=1\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7-2y\\21-6y-4y=1\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7-2y\\20=10y\end{matrix}\right.< =>\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
Vậy ...
b)\(\left\{{}\begin{matrix}y=7-2x\\4x-3\left(7-2x\right)=-1\end{matrix}\right.< =>\left\{{}\begin{matrix}y=7-2x\\4x-21+6x=-1\end{matrix}\right.< =>\left\{{}\begin{matrix}y=7-2x\\10x=20\end{matrix}\right.< =>\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)
a, \(\left\{{}\begin{matrix}x+2y=7\left(1\right)\\3x-4y=1\left(2\right)\end{matrix}\right.\)
Nhân cả 2 vế pt (1) với 3 ta được hệ phương trình
\(\left\{{}\begin{matrix}3x+6y=21\left(3\right)\\3x-4y=1\left(4\right)\end{matrix}\right.\)
Trừ 2 vế pt (3) cho pt (4)
=>10y=20
\(\Leftrightarrow y=2\) thay vào (1) ta có: x+4=7\(\Leftrightarrow x=3\)
Vậy nghiệm của hpt (x;y)=(3;2)
b,\(\left\{{}\begin{matrix}2x+y=7\left(1\right)\\4x-3y=-1\left(2\right)\end{matrix}\right.\)
Nhân 2 vế pt (1) vs 2 ta được
4x+2y=14(3)
Trừ 2 vế pt(3) cho pt(2)ta có
5y=15
\(\Leftrightarrow\)y=3 thay vào (1)
=>2x+3=7\(\Leftrightarrow x=2\)
Vậy nghiệm của hpt (x;y)=(2;3)
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.
\(\Leftrightarrow\left\{{}\begin{matrix}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x-6y=66\\-3x=-9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x^2+xy+3=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+x\left(1-x\right)+3=0\)
\(\Leftrightarrow x+3=0\Rightarrow x=-3\Rightarrow y=4\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{2x-5}{3}\\x^2-y^2=40\end{matrix}\right.\)
\(\Rightarrow x^2-\left(\frac{2x-5}{3}\right)^2-40=0\)
\(\Leftrightarrow9x^2-\left(4x^2-20x+25\right)-360=0\)
\(\Leftrightarrow5x^2+20x-385=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\Rightarrow y=3\\x=-11\Rightarrow y=-9\end{matrix}\right.\)
d.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{36-3x}{2}\\\left(x-2\right)\left(y-3\right)=18\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)\left(\frac{36-3x}{2}-3\right)=18\)
\(\Leftrightarrow\left(x-2\right)\left(10-x\right)=12\)
\(\Leftrightarrow-x^2+12x-32=0\Rightarrow\left[{}\begin{matrix}x=4\Rightarrow y=12\\x=8\Rightarrow y=6\end{matrix}\right.\)