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

Ta có: 

\(C=\sqrt{-x^2+6x}\) 

Mà: \(\sqrt{-x^2+6x}\ge0\) 

Dấu "=" xảy ra khi:

\(\sqrt{-x^2+6x}=0\)

\(\Leftrightarrow\sqrt{-x\left(x-6\right)}=0\)

\(\Leftrightarrow-x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: \(C_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

\(D=\sqrt{6x-2x^2}\)

Mà: \(\sqrt{6x-2x^2}\ge0\)

Dấu "=" xảy ra khi:

\(\sqrt{6x-2x^2}=0\)

\(\Leftrightarrow\sqrt{2x\left(3-x\right)}=0\)

\(\Leftrightarrow2x\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy: \(D_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(C=\sqrt{-x^2+6x}=\sqrt{9-\left(x^2-6x+9\right)}=\sqrt{9-\left(x-3\right)^2}\le\sqrt{9}=3\)

Dấu "=" xảy ra khi \(x=3\)

Vậy \(maxC=3\)

\(D=\sqrt{6x-2x^2}=\dfrac{1}{\sqrt{2}}\sqrt{12x-4x^2}=\dfrac{1}{\sqrt{2}}\sqrt{9-\left(4x^2-12x+9\right)}\)

\(=\dfrac{1}{\sqrt{2}}\sqrt{9-\left(2x-3\right)^2}\le\dfrac{1}{\sqrt{2}}.\sqrt{9}\)\(=\dfrac{3\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=\dfrac{3}{2}\)

Vậy \(maxD=\dfrac{3\sqrt{2}}{2}\)

toan violympic lop 9 la GTNN

\(B=\sqrt{x^2-6x+2y^2+4y+20}+\sqrt{x^2+2x+5}\)

\(=\sqrt{\left(x-3\right)^2+2\left(y+1\right)^2+9}+\sqrt{\left(x+1\right)^2+4}\ge\sqrt{9}+\sqrt{4}=5\)

tick nha

\(B=\sqrt{\left(3-x\right)^2+2\left(y+1\right)^2+3^2}+\sqrt{\left(x+2\right)^2+2^2}\ge\sqrt{\left(3-x+x+2\right)^2+\left(3+2\right)^2}=5\sqrt{2}\)

Bmin = \(5\sqrt{2}\) khi x=0 ; y =-1

B min nhé

Bài này đề sai nhé, trên ViOlympic đề sai. Nguyễn Văn Tiến giải sai rồi, bài này nếu đạt GTNN là 5 thì khi x bằng mấy ?

GTLN là 0
<=> x=3

\(A=5-\sqrt{x^2-6x+14}\)

   \(=5-\sqrt{\left(x^2-6x+9\right)+5}\)

  \(=5-\sqrt{\left(x-3\right)^2+5}\le5-\sqrt{5}\)

Dấu "=" <=> x-3=0

             <=> x=3

Vậy ....

Câu 2: 

\(C=-x+\sqrt{x}\)

\(=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)

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 \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2