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\(x^2-4xy+5y^2=16\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+y^2=16\)
\(\Leftrightarrow\left(x-2y\right)^2+y^2=16=4^2+0^2=0^2+4^2\)
\(TH1:\left\{{}\begin{matrix}\left(x-2y\right)^2=4^2\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4;x=-4\\y=0\end{matrix}\right.\)
\(TH2:\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\y^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)
\(xy+3x-y=38\)
\(\Leftrightarrow\left(xy-y\right)+\left(3x-3\right)=35\)
\(\Leftrightarrow y\left(x-1\right)+3\left(x-1\right)=35\)
\(\Leftrightarrow\left(x-1\right)\left(y+3\right)=35\)
Làm nốt
\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)
\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)
\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)
ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)
Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)
T i c k cho mình 1 cái nha mới bị trừ 50 đ
a ) \(x^2\left(x+3\right)+y^2\left(y+5\right)-\left(x+y\right)\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow x^3+3x^2+y^3+5y^2-\left(x^3+y^3\right)=0\)
\(\Leftrightarrow3x^2+5y^2=0\)
Do \(\left\{{}\begin{matrix}3x^2\ge0\forall x\\5y^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow3x^2+5y^2\ge0\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2=0\\5y^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy \(x=0;y=0\)
b )\(\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(-16\left(x^3-y\right)=32\)
\(\Leftrightarrow\left[\left(2x\right)^3-y^3\right]+\left[\left(2x\right)^3+y^3\right]-16x^3+16y=32\)
\(\Leftrightarrow8x^3-y^3+8x^3+y^3-16x^3+16y=32\)
\(\Leftrightarrow16y=32\)
\(\Leftrightarrow y=2\)
Vậy \(y=2\)
1. x2 + 2.\(\dfrac{1}{2}\)xy + \(\dfrac{1}{4}\)y2 + \(\dfrac{3}{4}\)y2 = 0
(=) (x+\(\dfrac{1}{2}\)y)2 +\(\dfrac{3}{4}\)y2 = 0
(=) \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}y\\y=0\end{matrix}\right.\) (=) x=y=0
2.
(x2 + 2xy +y2) + (x2 -2x +1) = 0
(=) (x+y)2 + (x-1)2 = 0
(=)\(\left\{{}\begin{matrix}x=-y\\x=1\end{matrix}\right.\) (=) \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
a) \(xy+x-y=2\)
\(\Leftrightarrow x\left(y+1\right)-\left(y+1\right)=1\)
\(\Leftrightarrow\left(x-1\right)\left(y+1\right)=1=1.1=\left(-1\right).\left(-1\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=y+1=1\\x-1=y+1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2;y=0\\x=0;y=-2\end{cases}}\)
b) \(x-2xy+y=0\)
\(\Leftrightarrow2x-4xy+2y=0\)
\(\Leftrightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)
\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=-1\)
Tương tự nha
c) \(x\left(x-2\right)-\left(2-x\right)y-2\left(x-2\right)=3\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)y-2\left(x-2\right)=3\)
\(\Leftrightarrow\left(x-2\right)\left(x+y-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)