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\(a,9x^2-6x+2\)
\(\left(3x-1\right)^2+1\ge1>0\)
vậy pt luôn dương
\(b,x^2+x+1\)
\(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
vậy pt luôn dương
\(c,2x^2+2x+1\)
\(\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\)
vậy pt luôn dương
Trả lời:
a, \(9x^2-6x+2=\left(3x\right)^2-2.3x.1+1+1=\left(3x-1\right)^2+1\ge1>0\forall0\)
Dấu "=" xảy ra khi 3x - 1 = 0 <=> x = 1/3
Vậy bt luôn dương với mọi x
b, \(x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
Dấu "=" xảy ra khi x + 1/2 = 0 <=> x = - 1/2
Vậy bt luôn dương với mọi x
c, \(2x^2+2x+1=2\left(x^2+x+\frac{1}{2}\right)=2\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)\)
\(=2\left[\left(x+\frac{1}{2}\right)^2+\frac{1}{4}\right]=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\forall x\)
Dấu "=" xảy ra khi x + 1/2 = - 1/2
Vậy bt luôn dương với mọi x
a. \(2x^2-4x+10=x^2-2x+1+x^2-2x+1+8=\left(x-1\right)^2+\left(x-1\right)^2+8=2\left(x-1\right)^2+8\)
Vì \(2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2+8\ge8\)
Vậy...
b. \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy..
c. \(2x^2-6x+5=x^2-4x+4+x^2-2x+1=\left(x-2\right)^2+\left(x-1\right)^2\)
Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}}\Rightarrow\left(x-2\right)^2+\left(x-1\right)^2\ge0\)
Vậy...
a, \(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1\ge1>0\forall x\)
Vậy ta có đpcm
b, \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
Vậy ta có đpcm
c, \(2x^2+2x+1=2\left(x^2+x+\frac{1}{4}-\frac{1}{4}\right)+1\)
\(=2\left(x+\frac{1}{2}\right)^2-\frac{1}{2}+1=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\forall x\)
Vậy ta có đpcm
\(a,-x^2+6x-16\)
\(=-x^2+3x+3x-9-5\)
\(=-x\left(x-3\right)+3\left(x-3\right)-5\)
\(=\left(3-x\right)\left(x-3\right)-5\)
\(=-\left(x-3\right)^2-5\le-5\)=>Luôn âm
\(c,-1+x-x^2\)
\(=-x^2+x-1\)
\(=-\left(x^2-x+\frac{1}{2}+\frac{1}{2}\right)\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\le\frac{-1}{2}\)=>Luôn âm
\(-5-\left(x-1\right)\left(x+2\right)=-5-\left(x^2+x-2\right)=-5-x^2-x+2\)
\(=-x^2-x-3=-\left(x+\frac{1}{2}\right)^2-\frac{11}{4}< 0,\forall x\inℝ\)
Với x > 0, áp dụng bất đẳng thức AM-GM ta có :
\(x+2021\ge2\sqrt{2021x}\Rightarrow\left(x+2021\right)^2\ge8084x\)
\(\Rightarrow\frac{1}{\left(x+2021\right)^2}\le\frac{1}{8084x}\Leftrightarrow\frac{x}{\left(x+2021\right)^2}\le\frac{1}{8084}\)
Dấu "=" xảy ra <=> x = 2021
Vậy ...
\(\frac{x}{\left(x+2021\right)^2}\left(x>0\right)\)
\(=\frac{1}{\frac{1}{x}\left(x+2021\right)^2}\)
\(=\frac{1}{\left(\frac{x+2021}{\sqrt{x}}\right)^2}\)
\(=\frac{1}{ \left(\sqrt{x}+\frac{2021}{\sqrt{x}}\right)^2}\)
Ta có :
\(\sqrt{x}+\frac{2021}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2021}{\sqrt{x}}}=2\sqrt{2021}\)
\(\rightarrow\left(\sqrt{x}+\frac{2021}{\sqrt{x}}\right)^2\ge4.2021=8084\)
\(\rightarrow\frac{1}{\left(\sqrt{x}+\frac{2021}{\sqrt{x}}\right)^2}\le\frac{1}{8084}\)
Dấu ''='' xảy ra \(\Leftrightarrow\sqrt{x}=\frac{2021}{\sqrt{x}}\Leftrightarrow x=2021\)
Vậy Max \(\left(\frac{x}{\left(x+2021\right)^2}\right)=\frac{1}{8084}\Leftrightarrow x=2021\)
a) Đặt \(P=\left(1-2x\right)\left(x-1\right)-5\)
\(P=x-1-2x^2+2x-5\)
\(P=-2x^2+3x-6\)
\(P=\dfrac{-\left(4x^2-12x+24\right)}{4}\)
\(P=\dfrac{-\left(4x^2-12x+9\right)}{4}-\dfrac{15}{4}\)
\(P=-\left(\dfrac{2x-3}{2}\right)^2-\dfrac{15}{4}\)
Ta thấy \(-\left(\dfrac{2x-3}{2}\right)^2\le0,\forall x\inℝ\Leftrightarrow P\le-\dfrac{15}{4}< 0,\forall x\inℝ\). Vậy ta có đpcm.
Câu b biểu thức đó sẽ bằng \(H=x^2-3x+7=x^2-2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}+7\) \(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\), do \(\left(x-\dfrac{3}{2}\right)^2\ge0\Leftrightarrow H\ge\dfrac{19}{4}>0\), vậy H sẽ luôn dương chứ không phải luôn âm.
câu a mình sửa lại nhé:
ở chỗ \(P=-2x^2+3x-6=-2\left(x^2-\dfrac{3}{2}x+3\right)\) r hướng làm tương tự