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\(2x^2+y^2+2xy-8x-6y+10=0\)
\(\Rightarrow2.\left(2x^2+y^2+2xy-8x-6y+10\right)=0\)
\(\Rightarrow4x^2+2y^2+4xy-16x-12y+20=0\)
\(\Rightarrow\left(4x^2+y^2+16+4xy-8y-16x\right)+\left(y^2-4y+4\right)=0\)
\(\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2=0\left(1\right)\)
Ta có: \(\hept{\begin{cases}\left(2x+y-4\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2\ge0\forall x;y\left(2\right)}\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x+y-4=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+y=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}2x+2=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
Chúc bạn học tốt.
\(5x^2+5y^2+8xy+2x-2y+2=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-2y+1\right)+4\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+4\left(x+y\right)^2=0\)
\(\Rightarrow x=-1;y=1\)
Khi đó:
\(M=\left(1-1\right)^{2010}+\left(2-1\right)^{2011}+\left(1-1\right)^{2012}\)
\(=1\)
\(Q=\left(x-3\right)\left(4x+5\right)+2019\)
\(=4x^2-7x-15+2019\)
\(=4x^2-7x+2004\)
\(=\left(2x-\frac{7}{4}\right)^2+\frac{32015}{16}\ge\frac{32015}{16}\forall x\)
Dấu "=" xảy ra<=>\(\left(2x-\frac{7}{4}\right)^2=0\Leftrightarrow2x=\frac{7}{4}\Leftrightarrow x=\frac{7}{8}\)
Ta có : \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
Khi đó : \(3x^{2018}=27^{673}=\left(3^3\right)^{673}=3^{2019}\)
\(\Leftrightarrow x^{2018}=3^{2018}\)
\(\Leftrightarrow\orbr{\begin{cases}x=y=z=3\\x=y=z=-3\end{cases}}\)
Đến đây tự tính A nha!
1a) A = \(x^2-4x+2023=\left(x-2\right)^2+2019\)
Ta luôn có: (x - 2)2 \(\ge\)0 \(\forall\)x
=> (x - 2)2 + 2019 \(\ge\)2019 \(\forall\)x
Hay A \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra khi : (x - 2)2 = 0 => x - 2 = 0 => x = 2
Nên Amin = 2019 khi x = 2
\(x^2+2y^2+2xy-2x-6y+5=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)-\left(2x+2y\right)+1+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)+1+\left(y-2\right)^2=0\)
\(\Leftrightarrow\left(x+y-1\right)^2+\left(y-2\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left(x+y-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Khi đó \(P=\dfrac{\left(-1\right)^2-7\cdot\left(-1\right)\cdot2+51}{-1-2}=-22\)