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a: Ta có: \(x^2-8x+20\)
\(=x^2-8x+16+4\)
\(=\left(x-4\right)^2+4>0\forall x\)
b: Ta có: \(-x^2+6x-19\)
\(=-\left(x^2-6x+19\right)\)
\(=-\left(x^2-6x+9+10\right)\)
\(=-\left(x-3\right)^2-10< 0\forall x\)
a) Ta có:
\(x^2+4x+5\)
\(=x^2+2.x.2+4+1\)
\(=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+1>0\forall x\)
\(\Rightarrow x^2+4x+5>0\forall x\)
b) Ta có:
\(x^2-x+1\)
\(=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
\(\Rightarrow x^2-x+1>0\forall x\)
c) Ta có:
\(12x-4x^2-10\)
\(=-\left(4x^2-12x+10\right)\)
\(=-\left[\left(2x\right)^2-2.2x.3+9+1\right]\)
\(=-\left(2x-3\right)^2-1\)
Vì \(-\left(2x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(2x-3\right)^2-1< 0\forall x\)
\(\Rightarrow12x-4x^2-10< -1\)
a, Sửa đề:
-x2-2x-2
=-(x2+2x+2)
=-(x2+2x+1+1)
=-[(x+1)2+1]<0\(\forall\)x
b, -x2-6x-11
=-(x2+6x+11)
=-(x2+2.x.3+32+2)
=-[(x+3)2+2]<0\(\forall\)x
Đúng tick nha,
a, -x - 2x - 2
= -(x+2x+1)-1
= -(x+1)2 -1
Có (x + 1)2 ≥0 ⇒- (x + 1) ≤ 0 ⇒ -(x + 1)2 - 1≤ -1
Do đó - x - 2x - 2 < 0 ∀ x
b, -x2 - 6x - 11
= -(x2 + 2.3.x+ 32)-2
= -(x+3)2 - 2
Có (x + 3)2 ≥0 ⇒- (x + 3) ≤ 0 ⇒ -(x + 3)2 - 2 ≤ -2
Do đó -x2 - 6x - 11 <0 ∀ x
\(A=9x^2-6x+2=\left(3x\right)^2-2.3x+1+1=\left(3x-1\right)^2+1>0\forall x\)
Vậy ta có đpcm
\(B=x^2-2xy+y^2+1=\left(x-y\right)^2+1>0\forall x;y\)
Vậy ta có đpcm
\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)
\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(đpcm\right)\)
\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)
\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)
\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)
a)
Đặt \(A=9x^2-6x+2\)
\(=\left(3x\right)^2-2.3x+1+1\)
\(=\left(3x+1\right)^2+1\)
Ta có: \(\left(3x+1\right)^2\ge0;\forall x\)
\(\Rightarrow\left(3x+1\right)^2+1\ge0+1;\forall x\)
Hay \(A\ge1>0;\forall x\)
Các phần khác tương tự cứ việc biến đổi thành hằng đẳng thức
\(a,9x^2-6x+2\)
\(=\left(3x\right)^2-2.3x.1+1^2+1\)
\(=\left(3x-1\right)^2+1\)
Vì\(\left(3x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(3x-1\right)^2+1\ge1>0\forall x\)
\(\Rightarrow9x^2-6x+2>0\forall x\)
\(b,x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
\(\Rightarrow x^2+x+1>0\forall x\)
Theo bài ra ta có:
\(x^2-x+1=x^2-\dfrac{1}{2}.2.x+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{2}=0=>x=\dfrac{1}{2}\)
Mà : \(\dfrac{3}{4}>0\)
\(=>x^2-x+1>0\)
CHÚC BẠN HỌC TỐT..........
\(x^2-x+1\\ = x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\\= \left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có:
\(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
Vậy \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\left(đpcm\right)\)
Ta có:
\(x^2-x+2=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\)
câu a mình nghĩ đề là\(x^2-x+1\)
b) \(x-x^2-2=-\left(x^2-x+2\right)=-[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}]\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{7}{4}\)<0 ∀\(x\)