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\(2x^2+y^2+4x-2y+3=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow2\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2\left(x+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
\(a,x^2+y^2-4x-2y+6\)
\(=\left(x^2-4x+4\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-2\right)^2+\left(y-1\right)^2+1\)
Ta có: \(\left(x-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-2\right)^2+\left(y-1\right)^2+1\ge1\forall x,y\)
Hay: \(x^2+y^2-4x-2y+6\ge1\)
\(b,x^2+4y^2+z^2-4x+4y-8z+25\)
\(=\left(x^2-4x+4\right)+\left(4y^2+4y+1\right)+\left(z^2-8z+16\right)+4\)
\(=\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2+4\)
Vì: \(\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2\ge0\forall x,y,z\)
\(\Rightarrow\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2+4\ge4\forall x,y,z\)
Hay: \(x^2+4y^2+z^2-4x+4y-8z+25\ge4\)
=.= hok tốt !!
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a)x2-2x-4y2-4y
=x2-2xy-2x+2xy-4y2-4y
=x(x-2y-2)+2y(x-2y-2)
=(x-2y-2)(x+2y)
c)x4+2x3-4x-4
=x4+2x3+2x2-2x2-4x-4
=x2(x2+2x+2)-2(x2+2x+2)
=(x2-2)(x2+2x+2)
Ta có: 4x2 - y2 + 4x + 4y - 3
= (4x2 - 4x + 1) - (y2 - 4y + 4)
= (2x - 1)2 - (y - 2)2
= (2x - 1 -y + 2)(2x - 1 + y - 2)
= (2x - y + 1)(2x + y - 3)
\(4x^2-y^2+4x+4y-3\)
\(=\left(4x^2+4x+1\right)-\left(y^2-4y+4\right)\)
\(=\left(2x+1\right)^2-\left(y-2\right)^2\)
\(=\left(2x+1+y-2\right)\left(2x+1-y+2\right)\)
\(=\left(2x+y-1\right)\left(2x-y+3\right)\)
x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử
= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung
= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung
\(x^2+4y^2-4x-4y+5=0\)
\(\Leftrightarrow\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(2y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)
\(x^2+4y^2-4x-4y+5=0\)
<=> \(\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)
<=> \(\left(x-2\right)^2+\left(2y-1\right)^2=0\)
<=> \(\hept{\begin{cases}x-2=0\\2y-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)
học tốt