sau khi bình pương và rút gọn biểu thức trong căn ta đc:

A²=2x²+10+2.\(\sqrt{\left(x^4\right)+6x^2+25}\)

vì x⁴+6x²+25>=25 với mọi x

nên .\(\sqrt{\left(x^4\right)+6x^2+25}\)>=5

=>2..\(\sqrt{\left(x^4\right)+6x^2+25}\)>=10

=>A²>=10+10=20

=>A>=\(\sqrt{20}\)

dấu = xảy ra khi x=0

vậy..

\(\sqrt{20}\)  khi x = 0

Ta có: \(\sqrt{x^2-2x+10}=\sqrt{x^2-2x+1+9}=\sqrt{\left(x-1\right)^2+9}\ge\sqrt{9}\ge3\)

          \(\sqrt{x^2+4x+5}=\sqrt{x^2+4x+4+1}=\sqrt{\left(x+2\right)^2+1}\ge\sqrt{1}\ge1\)

    \(\Rightarrow\)   \(\sqrt{x^2-2x+10}+\sqrt{x^2+4x+5}\ge1+3\ge4\)

Vậy GTNN của biểu thức là 4

a) \(A=\sqrt[]{x^2-2x+5}\)

\(\Leftrightarrow A=\sqrt[]{x^2-2x+1+4}\)

\(\Leftrightarrow A=\sqrt[]{\left(x+1\right)^2+4}\)

mà \(\left(x+1\right)^2\ge0,\forall x\in R\)

\(A=\sqrt[]{\left(x+1\right)^2+4}\ge\sqrt[]{4}=2\)

Dấu "=" xảy ra khi và chỉ khi \(x+1=0\Leftrightarrow x=-1\)

Vậy \(GTNN\left(A\right)=2\left(khi.x=-1\right)\)

b) \(B=5-\sqrt[]{x^2-6x+14}\)

\(\Leftrightarrow B=5-\sqrt[]{x^2-6x+9+5}\)

\(\Leftrightarrow B=5-\sqrt[]{\left(x-3\right)^2+5}\left(1\right)\)

Ta có : \(\left(x-3\right)^2\ge0,\forall x\in R\)

\(\Leftrightarrow\left(x-3\right)^2+5\ge5,\forall x\in R\)

\(\Leftrightarrow\sqrt[]{\left(x-3\right)^2+5}\ge\sqrt[]{5},\forall x\in R\)

\(\Leftrightarrow-\sqrt[]{\left(x-3\right)^2+5}\le-\sqrt[]{5},\forall x\in R\)

\(\Leftrightarrow B=5-\sqrt[]{\left(x-3\right)^2+5}\le5-\sqrt[]{5},\forall x\in R\)

Dấu "=" xả ra khi và chỉ khi \(x-3=0\Leftrightarrow x=3\)

Vậy \(GTLN\left(B\right)=5-\sqrt[]{5}\left(khi.x=3\right)\)

\(A=5-\sqrt{3-x^2+2x}\)

\(=5-\sqrt{-\left(x^2-2x-3\right)}\)

\(=5-\sqrt{-\left(x^2-2x+1-4\right)}\)

\(=5-\sqrt{-\left(x-1\right)^2+4}\)

\(A_{min}\Leftrightarrow\sqrt{-\left(x-1\right)^2+4}\)lớn nhất

Mà \(\left(x-1\right)^2\ge0\)\(\Rightarrow-\left(x-1\right)^2\le0\)

\(\Rightarrow-\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)=0\Rightarrow x=1\)

\(\Rightarrow A=5-\sqrt{4}=5-2=3\)

Vậy \(A_{min}=3\Leftrightarrow x=1\)

\(ĐKXĐ:3-x^2+2x\ge0\)

Ta co \(A=5-\sqrt{3-x^2+2x}=5-\sqrt{4-\left(x-1\right)^2}\ge5-\sqrt{4}=3\)

Dau "=" tai x = 1 (Tm ĐKXĐ)

Vay...

a . ta có : \(1\le1+\sqrt{2-x}\Rightarrow GTNN=1\)

\(-2\le\sqrt{x-3}-2\Rightarrow GTNN=-2\)

b. \(0\le\sqrt{4-x^2}\le2\)

\(\sqrt{2x^2-x+3}=\sqrt{2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{23}{8}}=\sqrt{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge\frac{\sqrt{46}}{4}\)

vậy \(GTNN=\frac{\sqrt{46}}{4}\)

ta có : \(0\le-x^2+2x+5=-\left(x-1\right)^2+6\le6\)

\(\Rightarrow1-\sqrt{6}\le1-\sqrt{-x^2+2x+5}\le1\)Vậy \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)

\(x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)

\(\sqrt{\left(x-1\right)^2+4}\ge2\)

\(\sqrt{x^2-2x+5}\ge2\)