Ta có: \(A-3\)

\(=\dfrac{x+5-3\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{x-3\sqrt{x}+2}{\sqrt{x}+1}\ge0\forall x\) thỏa mãn ĐKXĐ

hay A\(\ge3\)

Toán lớp 1 đây à ?

Giải:

\(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\ge0\)

\(\Leftrightarrow x^4\left(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\right)\ge0\)

\(\Leftrightarrow x^{12}-x^9-x^3+1\ge0\)

\(\Leftrightarrow x^9\left(x^3-1\right)-\left(x^3-1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^9-1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^3-1\right)\left(x^6+x^3+1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)^2\left(x^6+x^3+1\right)\ge0\) (luôn đúng)

Vậy ...

a) \(-\left(x^2-6x+10\right)=-\left(x^2-6x+9+1\right)=-\left[\left(x-3\right)^2+1\right]\le-1< 0\forall x\)

BĐT đúng

b) \(x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

BĐT đúng

c)Dấu "=" ko xảy ra???

\(=\left(4x^2+2.2x.y+y^2\right)+2\left(2x+y\right)+1+2\)

\(=\left(2x+y\right)^2+2.\left(2x+y\right).1+1+1\)

\(=\left(2x+y+1\right)^2+1\ge1>0\) (đpcm)

a. −x² + 6x - 10

= x² − 6x) − 10

= x² − 2.x.3 + 3² − 9) − 10

= x − 3)² + 9 − 10

= x − 3)² −1

Vì x − 3)² ≥ 0 ∀ x ⇒ x − 3)² ≤ 0 ⇒ x − 3)² −1 ≤ −1

Vậy x − 3)² −1 < 0 ⇒ −x² + 6x - 10 luôn âm với mọi x

b. x² + x + 1

= x² + 2.x.\(\frac{1}{2}\)+ (\(\frac{1}{2}\))²

)² + \(\frac{3}{4}\)

Vì (x + \(\frac{1}{2}\))² ≥ 0 ∀ x ⇒ (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) ≥ \(\frac{3}{4}\) ∀ x

Vậy (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) ≥ 0 hay x² + x + 1 > 0 ∀ x.

Xét hàm số \(f\left(x\right)=e^x-1-x\)  với \(x\ge0\)

Ta có : \(f'\left(x\right)=e^x-1\ge0\) với mọi  \(x\ge0\) 

và      : \(f'\left(x\right)=0\Leftrightarrow x=0\)

\(\Rightarrow f\left(x\right)\) đồng biến với  \(x\ge0\) nên với  \(x\ge0\Leftrightarrow f\left(x\right)\ge f\left(0\right)=0\)

a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)

Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)

b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)

hay \(P>0\forall x>0,x\ne1\)(đpcm)

P=(x²-1)²+(x-1)² lớn hơn = 0 với mọi x