Ta có:

\(x^2+1=x^2+xy+yz+zx\)

           \(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

\(=y\left(x+z\right)+z\left(x+z\right)\)

\(=\left(x+z\right)\left(y+z\right)\)

CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)

Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)

\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)

 Tương tự như thế, ta được

\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.

\(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)

\(=\sqrt{xy}+\sqrt{x}-\sqrt{y}\)

Ta có: \(P=\sqrt{xy}+\sqrt{x}-\sqrt{y}=2\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=3\)

\(\Rightarrow\left(\sqrt{x}-1,\sqrt{y}+1\right)=\left(1,3;3,1\right)\)

\(\Rightarrow\left(x,y\right)=\left(4,4;16,0\right)\)

\(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x\sqrt{x}+x-y+y\sqrt{y}-yx\sqrt{x}-xy\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x+y+\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)+2\sqrt{xy}-xy-1}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=1-\dfrac{\left(\sqrt{xy}-1\right)^2}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}=2\\ \Rightarrow\dfrac{\left(\sqrt{xy}-1\right)^2}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}=1\\ \Leftrightarrow\left(\sqrt{xy}-1\right)^2=\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)\\ \Leftrightarrow xy-2\sqrt{xy}+1=\sqrt{x}-\sqrt{y}+1-\sqrt{xy}\\ \Leftrightarrow\sqrt{x}-\sqrt{y}-xy+\sqrt{xy}=0\)

tự giải quyết tiếp nhá :)) h có việc :)) nếu còn ko bt thì mai làm nốt cho :))

2

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)

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

ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1

=> A ≥ 1

=> Min A =1 khi 1/3 ≤ x ≤ 2/3

câu 1) lm đơn giản chút nha . mà bài này đúng là \(x;y;z\) không âm phải không

vì đề bài bảo tính \(\Rightarrow\) giá trị của \(A\) là cố định

\(\Rightarrow\) chỉ cần tim \(x;y;z\) thỏa mãn điều kiện rồi thế vào là được

ta có : \(x=1;y=1;z=0\) thỏa mãn các điều kiện bài toán

thế vào \(A\) ta tính đc \(A=2\)

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)