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a/
\(\Leftrightarrow\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-6y+9\right)-19=0\)
\(\Leftrightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2=19\)
Do 19 không thể phân tích thành tổng của 2 số chính phương nên pt vô nghiệm
b/
\(\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Do x; y nguyên dương nên \(\left(2x+2y\right)^2>0\Rightarrow VT>0\)
Pt vô nghiệm
c/
\(\Leftrightarrow\left(x^2+4y^2+25-4xy+10x-20y+25\right)+\left(y^2-2y+1\right)+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)
Do x;y;z nguyên dương nên \(\left|x+y+z\right|>0\Rightarrow VT>0\)
Vậy pt vô nghiệm
d/
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Do x;y;z nguyên dương nên vế phái luôn dương
Pt vô nghiệm
a/ A = 2x2 + y2 - 2xy - 2x + 3
= (x2 - 2xy + y2) + (x2 - 2x + 1) + 2
= (x - y)2 + (x - 1)2 + 2\(\ge2\)
\(A=2x^2+y^2-2xy-2x+3\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\)
\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\)
Mà \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow A\ge2\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x-y=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=1\end{cases}}\)
Vậy Min A = 2 khi x=y=1
a, \(A_{\left(x\right)}=2x^2+2xy+y^2-2x+2y+2\)
\(=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(x^2-4x+4\right)-3\)
\(=\left(x+y+1\right)^2+\left(x-2\right)^2-3\ge-3\) hay \(A_{\left(x\right)}\ge-3\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy \(minA_{\left(x\right)}=-3\) khi x=-3; y=2
b, \(B_{\left(x\right)}=x^2-4xy+5y^2+10x-22y+28\)
\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\Leftrightarrow B_{\left(x\right)}\ge2\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
Vậy \(minB_{\left(x\right)}=2\Leftrightarrow x=-3;y=1\)
c, \(C_{\left(x\right)}=x^2-10xy+26y^2+14x-76y+59\)
\(=\left(x^2+25y^2+49-10xy+14x-70y\right)+\left(y^2-6y+9\right)+1\)
\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\Leftrightarrow C_{\left(x\right)}\ge1\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-5y+7\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-5y+7=0\\y-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)
Vậy \(minC_{\left(x\right)}=1\Leftrightarrow x=8;y=3\)
d, \(D_{\left(x\right)}=4x^2-4xy+2y^2-20x-4y+174\)
\(=\left(4x^2+y^2+25-4xy-20x+10y\right)+\left(y-14y+49\right)+74\)
\(=\left(2x-y-5\right)^2+\left(y-7\right)^2+74\ge74\Leftrightarrow D_{\left(x\right)}\ge74\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(2x-y-5\right)^2=0\\\left(y-7\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-y-5=0\\y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=7\end{matrix}\right.\)
Vậy \(minD_{\left(x\right)}=74\Leftrightarrow x=6;y=7\)
e, \(E_{\left(x\right)}=x^2-2x+y^2+4y+5\)
\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(minE_{\left(x\right)}=0\Leftrightarrow x=1;y=-2\)
bạn ơi! Sao cái chỗ A(x) =(x+y+1)2+(x-2)2-3 mà chuyển sang lại là -3 v
1) a) Đặt biểu thức là A
\(A=2x^2+4y^2-4xy-4x-4y+2017\)
\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)
\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)
\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)
Vậy: MinA=2008 khi x=-3; y=-2
3) a) \(A=\dfrac{1}{x^2+x+1}\)
\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)
Vậy MinA là \(\dfrac{4}{3}\) khi x=-0,5