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

\(\ge\sqrt{\left(3-x+x+1\right)^2+\left(3+2\right)^2}\text{ }\left(Mincopxki\right)\)

\(=\sqrt{41}\)

Đẳng thức xảy ra khi \(y+1=0\text{ và }\frac{3-x}{x+1}=\frac{3}{2}\Leftrightarrow y=-1;\text{ }x=\frac{3}{5}.\)

Vậy GTNN của A là \(\sqrt{41}\)

a: \(=\sqrt{x-3-2\sqrt{x-3}+3}\)

\(=\sqrt{x-3-2\sqrt{x-3}+1+2}=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}>=\sqrt{2}\)

Dấu = xảy ra khi x-3=1

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

em xin lỗi,em mới hok lớp 7 thôi

em cũng xin lỗi em học lớp 6

tôi không biết.

Ta có:

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

Áp dụng bđt Minkowski, ta có:

\(\Rightarrow A=\sqrt{\left(x-3\right)^2+2\left(y+1\right)^2}+\sqrt{\left(x+1\right)^2+3\left(y+1\right)^2}\)

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

\(A=\sqrt{4^2+\left(\sqrt{2}+\sqrt{3}\right)^2\left(y+1\right)^2}\ge\sqrt{4^2}=4\)

\(\Rightarrow A\ge4.Đ\text{TXR}\Leftrightarrow\orbr{\begin{cases}x=1;y=-1\\x=3;y=-1\end{cases}}\)

Dấu "=" xảy ra khi (x; y) = (3; -1)

\(A=\sqrt{2x^2-4x+3}+3\)

Ta có: \(2x^2-4x+3\)

\(=2\left(x^2-2x+\frac{3}{2}\right)\)

\(=2\left(x^2-2.x.1+1^2+\frac{1}{2}\right)\)

\(=2[\left(x-1\right)^2+\frac{1}{2}]\)

\(=2\left(x-1\right)^2+1\ge1\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}+3\ge3+\sqrt{1}=4\)

\(\Rightarrow MinA=4\Leftrightarrow x=1\)

Cần chứng minh bđt : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\Leftrightarrow\left(\left|a\right|+\left|b\right|\right)^2=\left(\left|a+b\right|\right)^2\)

\(\Leftrightarrow a^2+2\left|ab\right|+b^2\ge a^2+b^2+2ab\)

\(\Leftrightarrow\left|ab\right|\ge ab\) (luôn đúng)

Từ đó áp dụng ta được :

\(A\ge\sqrt{\left(x^2-6x+2y^2+4y+11\right)+\left(x^2+2x+3y^2+6y+4\right)}\)

\(\Leftrightarrow A\ge\sqrt{2x^2-4x+5y^2+10y+15}\)

\(\Leftrightarrow A\ge\sqrt{\left(2x^2-4x+2\right)+\left(5y^2+10y+5\right)+8}\)

\(\Leftrightarrow A\ge\sqrt{2\left(x-1\right)^2+5\left(y+1\right)^2+8}\ge\sqrt{8}=2\sqrt{2}\) có gtnn là \(2\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\)