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Ta có:
\(2010x^2+2011y^2-4020x+4022y+4021=0\)
\(\Leftrightarrow(2010x^2-4020x+2010)+(2011y^2+4022y+2011)=0\)
\(\Leftrightarrow2010(x^2-2x+1)+2011(y^2+2y+1)=0\)
\(\Leftrightarrow2010(x-1)^2+2011(y+1)^2=0\)
Ta thấy:
\(\left\{{}\begin{matrix}2010\left(x-1\right)^2\ge0\forall x\\2011\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow2010(x-1)^2+2011(y+1)^2\ge0\forall x,y\)
Mà \(\Leftrightarrow2010(x-1)^2+2011(y+1)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}2010\left(x-1\right)^2=0\\2011\left(y+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Có: \(8\left(a^2+b^2\right)=\left(2a+2b\right)^2\)
\(\Leftrightarrow8a^2+8b^2=4a^2+8ab+4b^2\)
\(\Leftrightarrow4a^2-8ab+4b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\Leftrightarrow a=b\)
=> đpcm
8(a2+b2) = (2a + 2b)2
=>8a2+8b2= 4a2 + 8ab + 4b
=> 4a2 + 4b2 = 8ab
=> 4a2 + 4b2 - 8ab = 0
=> (2a - 2b)2 =0
=> 2a - 2b = 0
=> 2(a-b)=0
=>a-b=0
=> a=b
ta có :
x2 = y2 (x + y4 + 2y2 )
<=> x2 -xy2 -y6 -2y4 =0
△= b2 -4ac
= y4 -4(-y6 -2y4 )
=9y4 +4y6 >=0 (mũ chẵn luôn luôn >=0)
=> pt có 2 nghiệm
x1 = (-b +√△)/2a = (y4 +√9y4 +4y6)/2
x2 = (-b - √△)/2a = (y4 - √9y4 +4y6)/2
dùng máy tính lập bảng để tìm nghiệm ta có : x= 12, y=2
\(2010x^2+2011y^2-4020x+4022y+4021=0\)
\(\Leftrightarrow\left(2010x^2-4020x+2010\right)+\left(2011y^2+4022y+2011\right)=0\)
\(\Leftrightarrow2010\left(x^2-2x+1\right)+2011\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow2010\left(x-1\right)^2+2011\left(y+1\right)^2=0\)
Vì \(2010\left(x-1\right)^2\ge0\forall x;2011\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2010\left(x-1\right)^2+2011\left(y+1\right)^2\ge0\)
Để \(2010\left(x-1\right)^2+2011\left(y+1\right)^2=0\Leftrightarrow\hept{\begin{cases}2010\left(x-1\right)^2=0\\2011\left(y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)
Vậy ..........