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\(-25x^2+5x-1=-\left(25x^2-5x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(5x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Ta có: với mọi số thực x
⇒ với mọi số thực x
⇒ với mọi số thực (ĐPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. Ta có : \(4x^2-6x+9=4x^2-6x+\dfrac{9}{4}+\dfrac{27}{4}\)
\(=\left[\left(2x\right)^2-6x+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{27}{4}\)
\(=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)
Vì \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\)
nên \(\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\forall x\)
b.Ta có : \(x^2+2y^2-2xy+y+1=\left(x^2+y^2-2xy\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\)
nên \(\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x;y\)
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\(=-\left(9x^2+2\cdot3\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{1}{4}-1=-\left(3x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\)
\(-9x^2+3x-1\)
\(=-9\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
\(=-9\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{1}{12}\right)\)
\(=-9\left(x-\dfrac{1}{6}\right)^2-\dfrac{3}{4}< 0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
x2 – 2xy + y2 + 1
= (x2 – 2xy + y2) + 1
= (x – y)2 + 1.
(x – y)2 ≥ 0 với mọi x, y ∈ R
⇒ x2 – 2xy + y2 + 1 = (x – y)2 + 1 ≥ 0 + 1 = 1 > 0 với mọi x, y ∈ R (ĐPCM).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x-x^2-1=-\left(x^2-x+\frac{1}{4}\right)-\frac{3}{4}=-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}< 0\)
Dấu "=" chỉ xảy ra khi:\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Vậy giá trị trên < 0 với mọi số thực x
\(x-x^2-1\\ =-\left(x^2-x+1\right)\\ =-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\\ \left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\in R\\ \Rightarrow-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]< 0\forall x\in R\\ \Leftrightarrow x-x^2-1< 0\forall x\in R\)
Vậy \(x-x^2-1< 0\forall x\in R\)
Ta có: \(x-x^2-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\right)\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\)
Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\in R\)
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le\dfrac{-3}{4}< 0\forall x\in R\)
-> ĐPCM.