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\(A=x-x^2=-x^2+x=-\left(x^2-x\right)=-\left(x^2-x+1-1\right)\)
\(=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}-1\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}-1\right]=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)
Dấu "=" xảy ra <=> \(\left(x-\frac{1}{2}\right)^2=0< =>x=\frac{1}{2}\)
Vậy MaxA=1/4 khi x=1/2
\(B=-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-2.x.3+9+2\right)=-\left[\left(x-3\right)^2+2\right]=-2-\left(x-3\right)^2\le-2\)
Dấu "=" xảy ra <=> x-3=0<=>x=3
Vậy maxB=-2 khi x=3
\(a^2+b^2+2a-2b-2ab=a^2-2ab+b^2+2\left(a-b\right)\)
\(=\left(a-b\right)^2+2\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+2\right)\)
\(4a^2-4b^2-4a+1=4a^2-4a+1-\left(2b\right)^2\)
\(=\left(2a-1\right)^2-\left(2b\right)^2\)
\(=\left(2a-1-2b\right)\left(2a-1+2b\right)\)
A= (4x2 + y2).[(2x)2 - y2] = (4x2 +y2)(4x2 - y2) = (4x2)2 _ (y2)2 = 16x4 - y4
\(A=x^2+4y^2-2xy+4x-10y+2020.\)
\(=\left(x^2-2xy+y^2\right)+\left(3y^2-6y+3\right)+\left(4x-4y\right)+2017\)
\(=\left(x-y\right)^2+3\left(y-1\right)^2+4\left(x-y\right)+2017\)
\(=\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]+3\left(y-1\right)^2+2013\)
\(=\left(x-y+2\right)^2+3\left(y-1\right)^2+2013\)
\(A_{min}=2013\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y+2=0\\y=1\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
\(B=8x^2+y^2-4xy-12x+2y+30\)
\(=\left(4x^2-4xy+y^2\right)+\left(4x^2-8x+4\right)-\left(4x-2y\right)+26\)
\(=\left(2x-y\right)^2+4\left(x-1\right)^2-2\left(2x-y\right)+26\)
\(=\left[\left(2x-y\right)^2-2\left(2x-y\right)+1\right]+4\left(x-1\right)^2+25\)
\(=\left(2x-y-1\right)^2+4\left(x-1\right)^2+25\)
\(\Rightarrow B_{min}=25\)\(\Leftrightarrow\hept{\begin{cases}\left(2x-y-1\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x-y-1=0\\x=1\end{cases}}\)\(\Leftrightarrow x=y=1\)
=x2-10x+25+76
=(x-5)2+76
Ta thấy:\(\left(x-5\right)^2+76\ge0+76=76\)
\(\Rightarrow A\ge76\)
Dấu "=" <=>x-5=0 =>x=5
Vậy...
=4a2+4a+1+1
=(2a+1)2+1
Ta thấy:\(\left(2a+1\right)^2+1\ge0+1=1\)
\(\Rightarrow B\ge1\)
Dấu "=" <=> 2a+1=0 <=>a=-1/2
=x2+4x+4-4
=(x+2)2-4
Ta thấy:\(\left(x+2\right)^2-4\ge0-4=-4\)
\(\Rightarrow C\ge-4\)
Dấu "=" <=> (x+2)=0 =>x=-2