\(\left\{{}\begin{matrix}x-y=3\\3x-4y=2\end{matrix}\right.\)
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30 tháng 4 2021 lúc 16:32
B4:Giải hệ pt:a)left{{}begin{matrix}4x+2y142x-2y4end{matrix}right.b)left{{}begin{matrix}2x-4y03x+2y8end{matrix}right.c)left{{}begin{matrix}2left(x+yright)+3left(x-yright)4left(x+yright)+2left(x-yright)5end{matrix}right.d)left{{}begin{matrix}dfrac{1}{x}+dfrac{1}{y}dfrac{1}{12}dfrac{8}{x}+dfrac{15}{y}1end{matrix}right.
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B4:Giải hệ pt:
a)\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;0\right\}\)
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b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;1\right\}\)
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d.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\) ta có hệ pt:
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}7b=\dfrac{1}{3}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+15\times\dfrac{1}{21}=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+\dfrac{5}{7}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a=\dfrac{2}{7}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{21}\\\dfrac{1}{x}=\dfrac{1}{28}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=21\\x=28\end{matrix}\right.\)
vậy hệ pt có ndn\(\left\{28;21\right\}\)
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giải hệ phương trình
1)left{{}begin{matrix}3x+4y112x-y-11end{matrix}right. 2)left{{}begin{matrix}3x+2y02x+y-1end{matrix}right. 3)left{{}begin{matrix}3x+dfrac{5}{2}y92x+dfrac{1}{3}y2end{matrix}right.
4)left{{}begin{matrix}-x+3y162x+y3end{matrix}right. 5)left{{}begin{matrix}dfrac{-3}{x-y}+dfrac{5}{2x+y}-2dfrac{4}{x-y}-dfrac{10}{2x+y}2end{matrix}right. 6)left{{}begin{matrix}dfrac{1}{x}-dfrac{1}{y}1dfrac{3}{x}+dfrac{4}{y}5end{matrix}right.
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giải hệ phương trình
1)\(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\) 2)\(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\) 3)\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}-x+3y=16\\2x+y=3\end{matrix}\right.\) 5)\(\left\{{}\begin{matrix}\dfrac{-3}{x-y}+\dfrac{5}{2x+y}=-2\\\dfrac{4}{x-y}-\dfrac{10}{2x+y}=2\end{matrix}\right.\) 6)\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
1. \(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\8x-4y=-44\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\11x=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-3\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=0\\4x+2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
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mọi người giải gúp mình với. Cần cực gấp a,left{{}begin{matrix}3x+2y-2-x+4y3end{matrix}right.b,left{{}begin{matrix}x+2y115x-3y3end{matrix}right.c,left{{}begin{matrix}10x-9y115x+21y36end{matrix}right.d,left{{}begin{matrix}2x+y3x+y2end{matrix}right.e,left{{}begin{matrix}x+y22x-3y9end{matrix}right.f,left{{}begin{matrix}x-2y115x+3y3end{matrix}right.g,left{{}begin{matrix}3x-y52x+3y18end{matrix}right.h,left{{}begin{matrix}5x+3y-73x-y-8end{matrix}right.
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mọi người giải gúp mình với. Cần cực gấp \(a,\left\{{}\begin{matrix}3x+2y=-2\\-x+4y=3\end{matrix}\right.b,\left\{{}\begin{matrix}x+2y=11\\5x-3y=3\end{matrix}\right.c,\left\{{}\begin{matrix}10x-9y=1\\15x+21y=36\end{matrix}\right.d,\left\{{}\begin{matrix}2x+y=3\\x+y=2\end{matrix}\right.e,\left\{{}\begin{matrix}x+y=2\\2x-3y=9\end{matrix}\right.f,\left\{{}\begin{matrix}x-2y=11\\5x+3y=3\end{matrix}\right.g,\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.h,\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\)
a, Ta có : \(\left\{{}\begin{matrix}3x+2y=-2\\-x+4y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3\left(4y-3\right)+2y=-2\\x=4y-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}12y-9+2y=-2\\x=4y-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}14y=7\\x=4y-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=\frac{1}{2}\\x=\frac{4.1}{2}-3=-1\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-1;\frac{1}{2}\right)\)
b, Ta có : \(\left\{{}\begin{matrix}x+2y=11\\5x-3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2y\\5\left(11-2y\right)-3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2y\\55-10y-3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2y\\-13y=-52\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2.4=3\\y=4\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)
c, Ta có : \(\left\{{}\begin{matrix}10x-9y=1\\15x+21y=36\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}30x-27y=3\\30x+42y=72\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-9y=1\\-69y=-69\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-9=1\\y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(1;1\right)\)
d, Ta có : \(\left\{{}\begin{matrix}2x+y=3\\x+y=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3-2x\\x+2-2x=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3-2x\\2-x=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3-2.0=3\\x=0\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(0;3\right)\)
e, Ta có : \(\left\{{}\begin{matrix}x+y=2\\2x-3y=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-y\\2\left(2-y\right)-3y=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-y\\4-2y-3y=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-y\\-5y=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2+1=3\\y=-1\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-1\right)\)
f, Ta có : \(\left\{{}\begin{matrix}x-2y=11\\5x+3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11+2y\\5\left(11+2y\right)+3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11+2y\\55+10y+3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11+2y\\13y=-52\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-4\right)\)
g, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=9-5=4\\x=3\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)
h, Ta có : \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x+3\left(3x+8\right)=-7\\y=3x+8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x+9x+24=-7\\y=3x+8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-\frac{31}{14}\\y=3.\left(-\frac{31}{14}\right)+8=\frac{19}{14}\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-\frac{31}{14};\frac{19}{14}\right)\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\left(x+y-3\right)^3=4y^3\left(x^2y^2+xy+\frac{45}{4}\right)\\x+4y-3=2xy^2\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+7y=\left(x+y\right)^2+x^2y+7x+4\\3x^2+y^2+8y+4=8x\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}2x+5y=xy+2\\x^2+4y+21=y^2+10x\end{matrix}\right.\)
a) left{{}begin{matrix}2x+5y33x-2y-8end{matrix}right.
b) left{{}begin{matrix}x+3y52x-5y-1end{matrix}right.
c) left{{}begin{matrix}3x-4y182x+y1end{matrix}right.
d)left{{}begin{matrix}x-2y+z122x-y+3z18-3x+3y+3z-9end{matrix}right.
e) left{{}begin{matrix}x-2y+4z13y-3z-77z14end{matrix}right.
f) left{{}begin{matrix}2x+y+3z2-x+4y-6z55x-y+3z-5end{matrix}right.
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a) \(\left\{{}\begin{matrix}2x+5y=3\\3x-2y=-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+3y=5\\2x-5y=-1\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3x-4y=18\\2x+y=1\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}x-2y+z=12\\2x-y+3z=18\\-3x+3y+3z=-9\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}x-2y+4z=13\\y-3z=-7\\7z=14\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}2x+y+3z=2\\-x+4y-6z=5\\5x-y+3z=-5\end{matrix}\right.\)
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
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Giải hệ phương trình
left{{}begin{matrix}4left(2x-y+3right)-3left(x-2y+3right)483left(3x-4y+3right)+4left(4x-2y-9right)48end{matrix}right.
left{{}begin{matrix}6left(x+yright)8+2x-3y5left(y-xright)5+3x+2yend{matrix}right.
left{{}begin{matrix}-2left(2x+1right)+1,53left(y-2right)-6x11,5-4left(3-xright)2y-left(5-xright)end{matrix}right.
left{{}begin{matrix}dfrac{8x-5y-3}{7}+dfrac{11y-4x-7}{5}12dfrac{9x+4y-13}{5}-dfrac{3left(x-2right)}{4}15end{matrix}right.
left{{}begin{matrix}2sqrt{3}x-sqrt{5}y2...
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Giải hệ phương trình
\(\left\{{}\begin{matrix}4\left(2x-y+3\right)-3\left(x-2y+3\right)=48\\3\left(3x-4y+3\right)+4\left(4x-2y-9\right)=48\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-2\left(2x+1\right)+1,5=3\left(y-2\right)-6x\\11,5-4\left(3-x\right)=2y-\left(5-x\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\\\dfrac{9x+4y-13}{5}-\dfrac{3\left(x-2\right)}{4}=15\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{5}y=2\sqrt{6}-\sqrt{15}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
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: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
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26 tháng 10 2021 lúc 17:19
Ghpt:
a) \(\left\{{}\begin{matrix}x^2+2y^2=2x-2xy+1\\3x^2+2xy-y^2=2x-y+5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4xy+4x^2+4y^2+\dfrac{3}{\left(x+y\right)^2}=7\\2x+\dfrac{1}{x+y}=3\end{matrix}\right.\)
giải hệ pt sau
aleft{{}begin{matrix}4x+y28x+3y5end{matrix}right. bleft{{}begin{matrix}3x_{ }-2y114x-5y3end{matrix}right. cleft{{}begin{matrix}4x+3y135x-3y_{ }-31end{matrix}right. Dleft{{}begin{matrix}7X+5Y193x+5y31end{matrix}right.
eleft{{}begin{matrix}7x-5y33x+10y62end{matrix}right. fleft{{}begin{matrix}2x+5y113x+2y11end{matrix}right. gleft{{}begin{matrix}x+3y4y-x+52x-y3x-...
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giải hệ pt sau
a\(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) b\(\left\{{}\begin{matrix}3x_{ }-2y=11\\4x-5y=3\end{matrix}\right.\) c\(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=_{ }-31\end{matrix}\right.\) D\(\left\{{}\begin{matrix}7X+5Y=19\\3x+5y=31\end{matrix}\right.\)
e\(\left\{{}\begin{matrix}7x-5y=3\\3x+10y=62\end{matrix}\right.\) f\(\left\{{}\begin{matrix}2x+5y=11\\3x+2y=11\end{matrix}\right.\) g\(\left\{{}\begin{matrix}x+3y=4y-x+5\\2x-y=3x-2\left(y+1\right)\end{matrix}\right.\)
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)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
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a)left{{}begin{matrix}2left|x-6right|+3left|y-1right|55left|x-6right|-4left|y+1right|1end{matrix}right.
b) left{{}begin{matrix}2left|x+yright|-left|x-yright|93left|x+yright|+2left|x-yright|+17end{matrix}right.
c)left{{}begin{matrix}4left|x+yright|+3left|x-yright|83left|x+yright|-5left|x-yright|6end{matrix}right.
d) left{{}begin{matrix}x^2-xy242x-3y1end{matrix}right.
e) left{{}begin{matrix}3x-4y+10xy3left(x+yright)-9end{matrix}right.
f) left{{}begin{matrix}2x+3y53x^2-y^2+2y4end{matrix}right.
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a)\(\left\{{}\begin{matrix}2\left|x-6\right|+3\left|y-1\right|=5\\5\left|x-6\right|-4\left|y+1\right|=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2\left|x+y\right|-\left|x-y\right|=9\\3\left|x+y\right|+2\left|x-y\right|+17\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}4\left|x+y\right|+3\left|x-y\right|=8\\3\left|x+y\right|-5\left|x-y\right|=6\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}x^2-xy=24\\2x-3y=1\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}3x-4y+1=0\\xy=3\left(x+y\right)-9\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}2x+3y=5\\3x^2-y^2+2y=4\end{matrix}\right.\)
a: Đặt |x-6|=a, |y+1|=b
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+3b=5\\5a-4b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
b: Đặt |x+y|=a, |x-y|=b
Theo đề, ta có: \(\left\{{}\begin{matrix}2a-b=19\\3a+2b=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{55}{7}\\b=-\dfrac{23}{7}\left(loại\right)\end{matrix}\right.\)
=>HPTVN
c: Đặt |x+y|=a, |x-y|=b
Theo đề ta có: \(\left\{{}\begin{matrix}4a+3b=8\\3a-5b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)
=>|x+y|=2 và x=y
=>|2x|=2 và x=y
=>x=y=1 hoặc x=y=-1
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giải hpt:
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+x+1=3y\\y^2+y+1=3x\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
ĐK: \(x>\dfrac{1}{4};y>-\dfrac{1}{4}\), đặt \(a=\dfrac{1}{\sqrt{12x-3}};b=\dfrac{1}{\sqrt{4y+1}}\)với a,b>0
khi đó, ta có hệ phương mới \(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}80a+40b=8\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}45a=3\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35.\dfrac{1}{15}+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\b=\dfrac{1}{15}\end{matrix}\right.\)
thay \(\dfrac{1}{\sqrt{12x-3}}=a\) hay \(\dfrac{1}{\sqrt{12x-3}}=\dfrac{1}{15}\Rightarrow\sqrt{12x-3}=15\Leftrightarrow12x-3=225\Leftrightarrow12x=228\Leftrightarrow x=19\left(TMĐK\right)\) thay \(\dfrac{1}{\sqrt{4y+1}}=b\) hay
\(\dfrac{1}{\sqrt{4y+1}}=\dfrac{1}{15}\Rightarrow\sqrt{4y+1}=15\Leftrightarrow4y+1=225\Leftrightarrow4y=224\Leftrightarrow y=56\left(TMĐK\right)\)
Vậy (x;y)=(9;56) là nghiệm duy nhất của hệ phương trình đã cho.
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b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
ĐK: x,y#0, khi đó \(\dfrac{1}{x}+\dfrac{1}{y}=4\Rightarrow x+y=4xy\)
Do đó \(x\left(1+4y\right)+y=2\Leftrightarrow x+4xy+y=2\Leftrightarrow x+x+y+y=2\Leftrightarrow2\left(x+y\right)=2\Leftrightarrow x+y=1\)
Mà \(4xy=x+y\Leftrightarrow4xy=1\Leftrightarrow xy=\dfrac{1}{4}\)
Vậy \(x+y=1;xy=\dfrac{1}{4}\)
Do đó x,y là nghiệm của phương trình:
\(t^2-t+\dfrac{1}{4}=0\)
\(\Delta=b^2-4ac=1-4.1.\dfrac{1}{4}=0\)
Phương trình có nghiêm kép \(x_1=x_2=-\dfrac{b}{2a}=-\dfrac{-1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x=y=\dfrac{1}{2}\left(nhận\right)\)
Vậy (x;y)=\(\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) là nghiệm duy nhất của hệ phương trình đã cho.
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c)\(\left\{{}\begin{matrix}x^2+x+1=3y\\y^2+y+1=3x\end{matrix}\right.\)
Trừ vế đối vế hai phương trình, ta được:
\(x^2+x+1-y^2-y-1=3y-3x\\ \Leftrightarrow x^2-y^2+4x+4y=0\\ \Leftrightarrow\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\\ \Leftrightarrow\left(x-y\right)\left(x+y+4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x+y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-x-4\end{matrix}\right.\)
+Với x=y thế vào \(x^2+x+1=3y\) ta được
\(x^2+x+1=3x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Do đó (x;y)=(1;1) là một nghiệm của hệ phương trình đã cho.
+Với y=-x-4 thế vào \(x^2+x+1=3y\) ta được
\(x^2+x+1=3\left(-x-4\right)\Leftrightarrow x^2+4x+13=0\Leftrightarrow\left(x+2\right)^2+9=0\)(*)
Mặt khác \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+9\ge0\Rightarrow\left(x+2\right)^2\ge-9>0\), do đó phương trình (*) vô nghiệm
Vậy (x;y)=(1;1) là nghiệm duy nhất của hệ phương trình đã cho.
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