giải phương trình sau
2x-y=3
x+2y=4
y=2x-3
x+2y=4
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2x^2 - 3x -2 = 0
<=>2x2+x-4x-2=0
<=>x.(2x+1)-2.(2x+1)=0
<=>(2x+1)(x-2)=0
<=>2x+1=0 hoặc x-2=0
<=>x=-1/2 hoặc x=2
x^2 +2y^2 - 2xy + 4y = -4
<=>x2+2y2-2xy+4y+4=0
<=>x2-2xy+y2+y2+4y+4=0
<=>(x-y)2+(y+2)2=0
<=>x-y=0 và y+2=0
*y+2=0
<=>x=-2
*x-y=0
<=>x=y=-2
1. 2x^2 - 3x - 2 = 0 <=> đen ta = 3^2 - 4x2x-2 = 25 > 0 <=> x1 = -0.5: x2= 2
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.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}2x-y=1\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}4x-2y=2\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}7x=7\\2x-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\2.1-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
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 \(\left(x;y\right)=\left(\dfrac{4}{17};-\dfrac{11}{17}\right)\)
Hệ \(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)\left(x-2y\right)^2=\left(x-2y\right)^2\\\sqrt{x-2y}+\sqrt{3x+2y}=4x-4\end{cases}.}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-2y\right)^2\left(x+y-1\right)=0\\\sqrt{x-2y}+\sqrt{3x+2y}=4x-4\end{cases}}\)
Đến đây thì đơn giản rồi, tự làm nhé
x+2(2x-3)=4
x+4x-6=4
5x=4+6
5x=10
x=2