Giải các hệ phương trình :
a) \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\);
b) \(\left\{{}\begin{matrix}\left(\sqrt{5}+2\right)x+y=3-\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\).
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a: \(\left\{{}\begin{matrix}\sqrt{5}x-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\sqrt{15}x-2\sqrt{3}\cdot y=2\sqrt{15}\left(\sqrt{3}-1\right)\\2\sqrt{15}x+15y=21\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2\sqrt{3}y-15y=2\sqrt{45}-2\sqrt{15}-21\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-2\sqrt{3}-15\right)=-15\sqrt{5}-2\sqrt{15}\\2\sqrt{3}\cdot x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{15\sqrt{5}+2\sqrt{15}}{2\sqrt{3}+15}=\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\2\sqrt{3}x=21-3\sqrt{5}\cdot\sqrt{5}=21-15=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\x=\dfrac{6}{2\sqrt{3}}=\sqrt{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y=19\\4,2x+10y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y+4,2x+10y=19,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12,7x=19,8\\2,1x+5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{198}{127}\\5y=0,4-2,1x=-\dfrac{365}{127}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{198}{127}\\y=-\dfrac{73}{127}\end{matrix}\right.\)
e.
\(\left\{{}\begin{matrix}2x-3y+5=0\\3x+5y-21=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y=-25\\9x+15y=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=38\\3x+5y=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{21-3x}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
f.
\(\left\{{}\begin{matrix}x-y\sqrt{2}=0\\2x\sqrt{2}+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y\sqrt{2}=0\\4x+y\sqrt{2}=5\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\sqrt{2}\\2x\sqrt{2}+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=5-2x\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=1\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}5x=-25\\3x-5y=-30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{3x+30}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=3\end{matrix}\right.\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}8x-6y=-10\\9x+6y=-24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}17x=-34\\9x+6y=-24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{-24-9x}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)
a: =>8x+2y=4 và 8x+3y=5
=>y=1 và 4x=2-1=1
=>x=1/4 và y=1
b: 3x-2y=11 và 4x-5y=3
=>12x-8y=44 và 12x-15y=9
=>7y=35 và 3x-2y=11
=>y=5 và 3x=11+2*y=11+2*5=21
=>x=7 và y=5
c: 5x-4y=3 và 2x+y=4
=>5x-4y=3 và 8x+4y=16
=>13x=19 và 2x+y=4
=>x=19/13 và y=4-2x=4-38/13=52/13-38/13=14/13
d: 3x-y=5 và 5x+2y=28
=>6x-2y=10 và 5x+2y=28
=>11x=38 và 3x-y=5
=>x=38/11 và y=3x-5=104/11-5=104/11-55/11=49/11
a: =>8x+2y=4 và 8x+3y=5
=>y=1 và 4x=2-1=1
=>x=1/4 và y=1
b: 3x-2y=11 và 4x-5y=3
=>12x-8y=44 và 12x-15y=9
=>7y=35 và 3x-2y=11
=>y=5 và 3x=11+2*y=11+2*5=21
=>x=7 và y=5
c: 5x-4y=3 và 2x+y=4
=>5x-4y=3 và 8x+4y=16
=>13x=19 và 2x+y=4
=>x=19/13 và y=4-2x=4-38/13=52/13-38/13=14/13
d: 3x-y=5 và 5x+2y=28
=>6x-2y=10 và 5x+2y=28
=>11x=38 và 3x-y=5
=>x=38/11 và y=3x-5=104/11-5=104/11-55/11=49/11
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
mấy bài này là ở lớp 9 học kì 2 dùng cộng đại số là nhanh nhất hoặc bấm máy tính
1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)
=> Hệ có vô số nghiệm.
3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
a)
\(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}17x-20y=38\\21x+50y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}85x-100y=190\\42x+100y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}127x=198\\21x+50y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\frac{198}{127}\\21.\frac{198}{127}+50y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\frac{198}{127}\\50y=4-\frac{4158}{127}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\frac{198}{127}\\50y=-\frac{3650}{127}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\frac{198}{127}\\y=-\frac{73}{127}\end{matrix}\right.\)
Vậy hệ pt có nghiệm duy nhất là (\(\left(\frac{198}{127};-\frac{73}{127}\right)\)
b)
\(\left\{{}\begin{matrix}\left(\sqrt{5}+2\right)x+y=3-\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2.\left(\sqrt{5}+2\right)x+2y=6-2\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2(\sqrt{5}+2)x=6+2\sqrt{5}-6-2\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2\left(\sqrt{5}+2\right)x=0\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\0+2y=6-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\frac{2\left(3-\sqrt{5}\right)}{52}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3-\sqrt{5}\end{matrix}\right.\)
Vậy hệ pt có nghiệm duy nhất là \(\left(0;3-\sqrt{5}\right)\)