`A=-10/(sqrtx+5)(x>=0)`

`x>=0=>sqrtx>=0`

`=>sqrtx+5>=5>0`

`=>10/(sqrtx+5)<=10/5=2`

`=>A>=-2`

Dấu "=" xảy ra khi `x=0`

Vậy GTNN `A=-2<=>x=0`

\(x+\dfrac{1}{x}=3\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\\ \Leftrightarrow x^3+\left(\dfrac{1}{x}\right)^3+3x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}+3\cdot3=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}=18\)

\(S=\dfrac{2x+2+x^2-2x+1}{x+1}=2+\dfrac{\left(x-1\right)^2}{x+1}\ge2\)

\(S_{min}=2\) khi \(x=1\)

\(B=\frac{x^2+17}{x^2+7}=\frac{x^2+7+10}{x^2+7}=1+\frac{10}{x^2+7}\)

\(B\)dat GTNN \(\Leftrightarrow\)\(\frac{10}{x^2+7}\)nho nhat\(\Leftrightarrow x^2+7\)lon nhat

Ma \(x^2+7>=7\)nen B khong co GTNN

\(P=\frac{\left(x+7\right)}{\sqrt{x}+3}=\)\(\frac{\left(x+3\sqrt{x}\right)-3\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}\)\(=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}\)

\(=\left(\sqrt{x}+3\right)+\frac{16}{\sqrt{x}+3}-6\)\(\ge8-6=2\)(AM-GM)

''='' <=> x = 1

Ta có :

\(A=x^2+3x+7=\left(x+1,5\right)^2+4,75\)

=> \(A_{Min}=4,75\Leftrightarrow x=-1,5\)

GTNN cua x la 7 nhaaa

Nham

GTNN cua A la 7 nhaaa

a) Ta có : \(M=x^2-4x+7\)

\(=x^2-4x+4+3\)

\(=\left(x-2\right)^2+3\)\(\ge3\forall x\in R\)(vì \(\left(x-2\right)^2\ge0\))

Vậy M_min = 3 khi x - 2 = 0 => x = 2

b)Ta có :  N = x² - x + 1 

\(=x^2-x+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\in R\)

Vậy N_min = \(\frac{3}{4}\) khi \(x=\frac{1}{2}\).

Đặt \(A=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)\)

\(=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)

\(=\left(x^2-7x\right)^2-100\ge-100\)

Dấu " = " khi \(x^2-7x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

Vậy \(MIN_A=-100\) khi x = 0 hoặc x = 7