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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)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
1) hpt \(\Leftrightarrow\left\{{}\begin{matrix}x+4y=2\\6x+4y=8\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2-x}{4}\\5x=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=\dfrac{6}{5}\end{matrix}\right.\)
Kl: x=6/5 và y=1/5
2) hpt \(\Leftrightarrow\left\{{}\begin{matrix}-2x-2y=4\\-2x-4y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2-y\\2y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Kl...
3) hpt \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=2\\2x-3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+3y}{2}\\0=3\left(vô-lý\right)\end{matrix}\right.\)
kl: hpt vn
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
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
a) \(\left\{{}\begin{matrix}x-y=3\left(1\right)\Rightarrow y=x-3\left(3\right)\\3x-4y=2\left(2\right)\end{matrix}\right.\)
thay (3) vào (2)\(\Rightarrow3x-4\left(x-3\right)=2\)
\(\Leftrightarrow3x-4x+12=2\)
\(\Leftrightarrow-x=-10\Leftrightarrow x=10\)
thay x=10 vào (3)\(\Rightarrow y=10-3=7\)
Nghiệm của hệ \(\left\{10;7\right\}\)
b)\(\left\{{}\begin{matrix}7x-3y=5\left(1\right)\\4x+y=2\left(2\right)\Rightarrow y=2-4x\left(3\right)\end{matrix}\right.\)
thay (3) vào (1)\(\Rightarrow7x-3\left(2-4x\right)=5\)
\(\Leftrightarrow7x-6+12x=5\)
\(\Leftrightarrow19x=11\Leftrightarrow x=\dfrac{11}{19}\)
thay \(x=\dfrac{11}{19}vào\left(3\right)\)\(\Rightarrow y=2-4\dfrac{11}{19}=-\dfrac{6}{19}\)
nghiệm của hệ \(\left\{\dfrac{11}{19};\dfrac{-6}{19}\right\}\)
c)\(\left\{{}\begin{matrix}x+3y=-2\left(1\right)\Rightarrow x=-2-3y\left(3\right)\\5x-4y=1\left(2\right)\end{matrix}\right.\)
thay (3) vào (2)\(\Rightarrow5\left(-2-3y\right)-4y=1\)
\(\Leftrightarrow-10-15y-4y=1\)
\(\Leftrightarrow-19y=11\Leftrightarrow y=\dfrac{-11}{19}\)
thay \(y=\dfrac{-11}{19}vào\left(3\right)\Rightarrow x=-2-3\left(\dfrac{-11}{19}\right)=\dfrac{-5}{19}\)nghiệm của hệ \(\left\{\dfrac{-5}{9};\dfrac{-11}{19}\right\}\)
c)\(\left\{{}\begin{matrix}x+3y=-2\left(1\right)\Rightarrow x=-2-3y\left(3\right)\\5x-4y=1\left(2\right)\end{matrix}\right.\)
thay (3) vào (2)\(\Rightarrow5\left(-2-3y\right)-4y=1\)
\(\Leftrightarrow-10-15y-4y=1\)
\(\Leftrightarrow-19y=11\Leftrightarrow y=\dfrac{-11}{19}\)
thay \(y=\dfrac{-11}{19}vào\left(3\right)\Rightarrow x=-2-3\left(\dfrac{-11}{19}\right)=\dfrac{-5}{19}\)
nghiệm của hệ\(\left\{\dfrac{-5}{19};\dfrac{-11}{19}\right\}\)
CHÚC BẠN HỌC TỐT !
-có người nhờ t làm
\(\left\{{}\begin{matrix}x-y=3\\3x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-3y=9\left(1\right)\\3x-4y=2\left(2\right)\end{matrix}\right.\) lấy (1)-(2) tìm được x;sau đó dễ dàng có y
\(\left\{{}\begin{matrix}7x-3y=5\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}28x-12y=20\left(1\right)\\28x+7y=14\left(2\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+3y=-2\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\left(1\right)\\5x-4y=11\left(2\right)\end{matrix}\right.\)
Gt: Nhân sao cho cả 2 pt xuất hiện chung 1 thừa số,trừ đi chỉ còn 1 x or y
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: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
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) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
Câu a: Thế y=5-2x rồi giải pt bậc2
Câu b : từ pt thứ 2, tương đương (x-3)(y-3)=0, xét 2 TH rồi thế vào pt thứ 1
Câu c: từ pt 1 suy ra 2x = 2-3y
Nhân 2 vào pt 2 rồi thế vào
a, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3.3-5=4\\x=\frac{33}{11}=3\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 3;4 )
b, Làm tương tự a
c, Ta có : \(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{15}{x-y+2}+\frac{10}{x+y-1}=20\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{29}{x-y+2}=29\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-y+2=1\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{3}{y-1-y+2}+\frac{2}{y-1+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\3+\frac{2}{2y-2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{2}{2y-2}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\2y-2=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-1=1\\y=2\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 1;2 )
\(\left\{{}\begin{matrix}x-y=4\\3x+4y=19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-3y=12\\3x+4y=19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\3x-3y-3x-4y=12-19=-7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+4\\-7y=-7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)
\(Vậy:\left(x,y\right)=\left(5,1\right)\)
\(\left\{{}\begin{matrix}x-y=4\\3x+4y=19\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}4x-4y=16\\3x+4y=19\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}7x=35\\x-y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)