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)
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
a: Ta có: \(x^2-8x+20\)
\(=x^2-8x+16+4\)
\(=\left(x-4\right)^2+4>0\forall x\)
b: Ta có: \(-x^2+6x-19\)
\(=-\left(x^2-6x+19\right)\)
\(=-\left(x^2-6x+9+10\right)\)
\(=-\left(x-3\right)^2-10< 0\forall x\)
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: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)
Vẽ đồ thị các hàm số :
a ) \(y=\hept{\begin{cases}2x\forall x\ge0\\x\forall x< 0\end{cases}}\)
b ) \(y=\hept{\begin{cases}2x\forall x\ge0\\-\frac{1}{2}x\forall x< 0\end{cases}}\)
Tìm GTLN của biểu thức: M= - 9x2+6x-3>0,\(\forall\)x
`M=-9x^2+6x-3`
`M=-(9x^2-6x+3)`
`M=-(9x^2-6x+1+2)`
`M=-(3x-1)^2-2`
Vì `-(3x-1)^2 <= 0 AA x`
`<=>-(3x-1)^2-2 <= -2 AA x`
Hay `M <= -2 AA x`
Dấu "`=`" xảy ra `<=>(3x-1)^2=0<=>3x-1=0<=>x=1/3`
Vậy `GTLN` của `M` là `-2` khi `x=1/3`
\(M=-9x^2+6x-3\)
\(M=-\left(9x^2-6x+3\right)\)
\(M=-\left[\left(3x-1\right)^2+2\right]\)
\(M=-\left(3x-1\right)^2-2\)
\(\Rightarrow Max_M=-2\) khi \(3x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
`-9x^2 + 6x - 3`.
`= -(3x - 1)^2 - 2`.
Vì `(3x-1)^2 >=0 => -(3x-1)^2 <=0 => -(3x-1)^2 - 2 <= -2`
Dấu bằng xảy ra `<=> 3x - 1 = 0 => x = 1/3`.
Vậy `Max_M = -2 <=> x = 1/3`.
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
chứng minh rằng :
a, x+2y+\(\dfrac{25}{x}\)+\(\dfrac{27}{y^2}\)\(\ge\) 19 ( \(\forall\)x,y \(\)> 0 )
b, \(x+\dfrac{1}{\left(x-y\right)y}\ge3\) ( \(\forall\)x>y>0 )
c,\(\dfrac{x}{2}+\dfrac{16}{x-2}\ge13\left(\forall x>2\right)\)
d, \(a+\dfrac{1}{a^2}\ge\dfrac{9}{4}\left(\forall x\ge2\right)\)
e, a+\(\dfrac{1}{a\left(a-b\right)^2}\ge2\sqrt{2}\) ( \(\forall x>y\ge0\))
f, \(\dfrac{2a^3+1}{4b\left(a-b\right)}\ge3[\forall a\ge\dfrac{1}{2};\dfrac{a}{b}>1]\)
g, x+\(\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}\ge3\left(\forall x>y\ge0\right)\)
h, \(2a^4+\dfrac{1}{1+a^2}\ge3a^2-1\)
Cho f(x)=2x+1. Khẳng định nào sau đây là sai:
A.f(x)>0,∀x>\(\dfrac{-1}{2}\)
B.f(x)>0,∀x<\(\dfrac{1}{2}\)
C.f(x)>0,∀x>2
D.f(x)>0,∀x>0
Xét tính đúng sai của mỗi mệnh đề sau:
∀x,y∈R,x^2+xy+y^2≥0
∃x,y∈R,x^2+y^2+xy<0
y=-x-6x có GTLN bằng 9
∀n∈N,(n^2+7n+12)⋮2
