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1) Biến đổi vế trái , ta có :
\(x^2+xy+y^2+1\)
\(\Leftrightarrow x^2+xy+\frac{1}{4}y^2+\frac{3}{4}y^2+1\)
\(\Leftrightarrow\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\left(đpcm\right)\)
x2 + xy + y2 + 1
\(=\left[x^2+2\cdot x\cdot\frac{y}{2}+\left(\frac{y}{2}\right)^2\right]+\frac{3y^2}{4}+1\)
\(=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\ge1>0\forall x,y\left(đpcm\right)\)
\(4x-x^2\)
\(=-\left(x^2-4x+4\right)+4\)
\(=-\left(x-2\right)^2+4\le4\forall x\)
\(-x^2+4x-10\)
\(=-\left(x^2-4x+4\right)-6\)
\(=-\left(x-2\right)^2-6\le-6< 0\forall x\left(đpcm\right)\)
\(Q=x^2+y^2+xy+x+y+10\)
\(=\left(x^2+xy+x\right)+y^2+y+10\)
\(=x^2+x\left(y+1\right)+y^2+y+10\)
\(=x^2+2.x.\frac{y+1}{2}+\left(\frac{y+1}{2}\right)^2+y^2+y-\left(\frac{y+1}{2}\right)^2+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{\left(y+1\right)^2}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{y^2+2y+1}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{1}{4}y^2-\frac{1}{2}y-\frac{1}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}y^2+\frac{1}{2}y+\frac{39}{4}\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+\frac{2}{3}y+13\right)=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+2.y.\frac{2}{6}+\frac{4}{36}-\frac{4}{36}+13\right)\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left[\left(y+\frac{2}{6}\right)^2+\frac{116}{9}\right]=\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\)
Vì \(\left(\frac{2x+y+1}{2}\right)^2\ge0;\frac{3}{4}\left(y+\frac{2}{6}\right)^2\ge0=>\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\ge\frac{29}{3}>0\) (với mọi x;y)
Vậy biểu thức Q luôn dương với mọi giá trị của biến
=>4Q=4x2+4xy+4y2+4x+4y+40
=4x2+4x(y+1)+(y+1)2+4y2-y2+4y-2y+40-1
=(2x+y+1)2+3y2+2y+39
\(=\left(2x+y+1\right)^2+\left(\sqrt{3}y+\frac{\sqrt{3}}{3}\right)^2+\frac{116}{3}\)
\(\Rightarrow Q=\left(\frac{2x+y+1}{2}\right)^2+\left(\frac{\sqrt{3}y+\frac{\sqrt{3}}{3}}{2}\right)^2+\frac{29}{3}>0\)
=>đpcm
1. Đặt \(t=x^2,t\ge0\)
\(3x^4+4x^2-2\ge3.0+4.0-2=-2\)
=> MIN = -2 khi x = 0
2. \(\left(x^2+2\right)\left(x+1\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+2=0\\x+1=0\end{array}\right.\)
Vì \(x^2+2\ge2>0\) => Vô nghiệm
Vậy x+1 = 0 => x = -1
3. Kết quả là 10
4. Ko rõ đề
Ta có A = -x2 + 4x - 6 - y2 - 2y
= -(x2 - 4x + 4) - (y2 + 2y + 1) - 1
= -(x - 2)2 - (y + 1)2 - 1 \(\le-1< 0\)
=> A < 0 với mọi x ; y
A = -x2 + 4x - 6 - y2 - 2y
= -( x2 - 4x + 4 ) - ( y2 + 2y + 1 ) - 1
= -( x - 2 )2 - ( y - 1 )2 - 1 ≤ -1 < 0 ∀ x, y
=> đpcm