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a,\(-\left(x^2-3x+4\right)\)
\(-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\Leftrightarrow-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)(luôn âm)
b\(-2\left(x^2-5x+\frac{15}{2}\right)\)
\(-2\left[\left(x-\frac{5}{2}\right)^2+\frac{5}{4}\right]\)
\(-2\left(x-\frac{5}{4}\right)^2-\frac{5}{2}\le-\frac{5}{2}\)(luôn âm)
c,\(-\left[\left(4x^2-4x+1\right)+\left(2y^2-6y+5\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y^2-3y+\frac{5}{2}\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2+\frac{1}{4}\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2\right]-\frac{1}{4}\le-\frac{1}{4}\)(luôn âm)
Ta có : x2 + 2xy + 2y2 + 2y + 1
= (x2 + 2xy + y2) + (y2 + 2y + 1)
= (x + y)2 + (y + 1)2
Vì : (x + y)2 \(\ge0\forall x\) ; (y + 1)2 \(\ge0\forall x\)
Nên : (x + y)2 + (y + 1)2 \(\ge0\forall x\)
Vậy (x + y)2 + (y + 1)2 không âm
2. Ta có: P = 2x2 + y2 - 4x - 4y + 10
P = 2(x2 - 2x + 1) + (y2 - 4y + 4) + 4
P = 2(x - 1)2 + (y - 2)2 + 4 \(\ge\)4 \(\forall\)x;y
=> P luôn dương với mọi biến x;y
3 Ta có:
(2n + 1)(n2 - 3n - 1) - 2n3 + 1
= 2n3 - 6n2 - 2n + n2 - 3n - 1 - 2n3 + 1
= -5n2 - 5n = -5n(n + 1) \(⋮\)5 \(\forall\)n \(\in\)Z
Câu 1:
\(a,P=x^2-2x+5=\left(x^2-2x+1\right)+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Vậy Min \(P=4\) khi \(x-1=0\Rightarrow x=1\)
\(b,Q=2x^2-6x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\)
\(=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)
Vậy \(MinQ=-\dfrac{9}{2}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(c,M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+9y+9\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Vậy Min \(M=\dfrac{3}{4}\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)