CHỨNG MINH :
a/ \(x^2-8x+20>0\forall x\)
b/ \(6x-x^2-19< 0\forall x\)
c/ \(3x^2+y^2-2xy+4x+20>0\forall x,y\)
d/ \(5x^2+10y^2-6xy-4x-2y+3>0\forall x,y\)
AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE NHA
CHỨNG MINH :
a/ \(3x^2+y^2-2xy+4x+20\forall x,y\)
b/ \(5x^2+10y^2-6xy-4x-2y+3\forall x,y\)
AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE NHA
a) \(3x^2+y^2-2xy+4x+20=\left(x^2-2xy+y^2\right)+2\left(x^2+2x+1\right)+18=\left(x-y\right)^2+2\left(x+1\right)^2+18\ge18>0\forall x,y\)
\(ĐTXR\Leftrightarrow x=y=-1\)
CMR:
a,\(x^2+5y^2+2x-4xy-10y+10>0\forall x,y\)
b,\(5x^2+10y^2-6xy-4x-2y+3>0\forall x,y\)
chứng minh rằng
a) 9x2-6x+2>0 \(\forall x \)
b)x2+x+1>0 \(\forall x \)
c) 25x2-20x+7>0 \(\forall x \)
d)9x2-6xy+2y2+1>0 \(\forall x ,y\)
e) x2-xy+y2 \(\ge0\forall x,y\)
\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)
\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(đpcm\right)\)
\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)
\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)
\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)
Cách khác câu e:
\(x^2-xy+y^2=x^2-2x.\frac{y}{2}+\frac{y^2}{4}+\frac{3y^2}{4}=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge0\forall xy\) (đpcm)
chúng minh rằng
a) 9x2-6x+2>0 \(\forall x \)
b)x2+x+1>0 \(\forall x \)
c) 25x2-20x+7>0 \(\forall x \)
d)9x2-6xy+2y2+1>0 \(\forall x ,y\)
e) x2-xy+y2 \(\ge0\forall x,y\)
hãy giúp mình nhé
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\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
\(\Rightarrow x^2+x+1>0\forall x\)
À xin lỗi sửa sai chút là \(\left(3x-1\right)^2\)nhé
Chứng minh
A = 9x^2 - 6x + 2 >0, \(\forall\)x
B = x^2 - 2xy + y^2 + 1 > 0, \(\forall\)x,y
\(A=9x^2-6x+2=\left(3x\right)^2-2.3x+1+1=\left(3x-1\right)^2+1>0\forall x\)
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
Trả lời:
\(A=9x^2-6x+2=\left(3x\right)^2-2.3x.1+1+1=\left(3x-1\right)^2+1\ge1>0\forall x\)
Vậy A > 0 với mọi x
\(B=x^2-2xy+y^2+1=\left(x-y\right)^2+1\ge1>0\forall x;y\)
Vậy B > 0 với mọi x;y
Bài 1 : cm
a) \(^{x^2}\)-2xy +\(^{y^2}\)+ 1 > 0 \(\forall\)xy
b) x-\(x^2\)-1<0 \(\forall\)x
c) 9\(x^2\)+12x+10 >0
d) 3\(^{x^2}\)-x+ 1>0
Giúp mk ,mk cần gấp
ai nhanh mk tick cho !!
a)\(x^2-2xy+y^2+1=\left(x+y\right)^2+1\ge1>0\)
b)\(x-x^2-1=-\left(x^2-x+\frac{1}{4}\right)^2-\frac{3}{4}\le-\frac{3}{4}< 0\)
c)\(9x^2+12x+10=\left(9x^2+12x+4\right)+6=\left(3x+2\right)^2+6\ge6>0\)
d)\(3x^2-x+1=2x^2+\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=2x^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0`\)
a) Giải phương trình: \(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
b) Cho \(0< x< y\le3\) và \(2xy\le3x+y\forall x,y\in R\). Chứng minh rằng: \(x^2+y^2\le10\)
Chứng minh BĐT:
a) x2 + x + 1 > 0 ∀ x
b) x - \(\sqrt{x}\) + 1 > 0 ∀ x
c) x2 - xy + y2 > 0 ∀ xy , x; y ≠0
d) x2 + x\(\sqrt{2}\) + 1 > 0 ∀ x
e) ( x + y + z )2 ≤ 3( x2 + y2 + z2) ∀ xyz
a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)
Chứng minh rằng
a) \(x^2+4x+5>0\forall x\)
b)\(x^2-x+1>0\forall x\)
c)\(12x-4x^2-10< -1\forall x\)
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 \(-\left(2x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(2x-3\right)^2-1< 0\forall x\)
\(\Rightarrow12x-4x^2-10< -1\)