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Có xy + yz + zx = 1
=> 1 + x2 = x2 + xy + yz + zx
1 + x2 = (x + y)(y + z)
Tương tự ta có:
1 + y2 = (y + x)(y + z)
1 + z2 = (z + x)(z + y)
Thay vào P, ta được:
\(P=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(P=xy+yz+zx+xy+yz+zx\)
\(P=2\left(xy+yz+zx\right)=2\)
Vậy P = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2+y^2+xy=\left(x+y\right)^2-2xy+xy\\ A=1-xy\)
Mà \(x+y=1\Leftrightarrow x=1-y\)
\(\Leftrightarrow A=1-\left(1-y\right)y=1-y+y^2=\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ A=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ A_{min}=\dfrac{3}{4}\Leftrightarrow x=y=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=a\)
\(\Rightarrow x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+\left(1+x^2\right)\left(1+y^2\right)=a^2\)
\(\Rightarrow x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2.x\sqrt{1+y^2}.y\sqrt{1+x^2}+1=a^2\)
\(\Rightarrow\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2+1=a^2\)
\(\Rightarrow E^2+1=a^2\)
\(\Rightarrow E=\pm\sqrt{a^2-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E^2=x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(y^2+1\right)\left(x^2+1\right)}\)
\(=2\left(xy\right)^2+x^2+y^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}\)
\(a^2=\left(xy\right)^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+\left(x^2+1\right)\left(y^2+1\right)\)
\(=2\left(xy\right)^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+x^2+y^2+1\)
\(\Rightarrow E^2=a^2-1\Rightarrow E=\sqrt{a^2-1}\)
\(E=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)
\(\Leftrightarrow E^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+y^2\right)\left(1+x^2\right)}\)
\(=2x^2y^2+x^2+y^2+2xy\left(a-xy\right)\)
\(=2x^2y^2+x^2+y^2+2xya-2x^2y^2\)
\(=x^2+y^2+2xya\)
\(=\left(2xy\right)2+a=a^2+a=E^2\)
\(E=\sqrt{a^2+a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\dfrac{1}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{y}-\sqrt{x}}\right):\dfrac{2\sqrt{xy}}{x-y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}}{x-y}:\dfrac{2\sqrt{xy}}{x-y}=\dfrac{-2\sqrt{y}}{2\sqrt{xy}}=\dfrac{-1}{\sqrt{x}}=\dfrac{-\sqrt{x}}{x}\)
b, Ta có \(A=\dfrac{-1}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=-1\left(voli\right)\)
Vậy pt vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{x}{1-x}+\frac{y}{1-y}=1\)
\(\Leftrightarrow\frac{x+y-2xy}{\left(x-1\right)\left(y-1\right)}=1\)
\(\Rightarrow x+y-2xy=xy-x-y+1\)
\(\Rightarrow2\left(x+y\right)-1=3xy\)
Lại có: \(P=x+y+\sqrt{x^2-xy+y^2}\)
\(=x+y+\sqrt{\left(x+y\right)^2-3xy}\)
\(=x+y+\sqrt{\left(x+y\right)^2-2\left(x+y\right)+1}\)
\(=x+y+\sqrt{\left(x+y-1\right)^2}\)
Mặt khác: \(\frac{x}{1-x}+\frac{y}{1-y}=1\); \(0< x;y< 1\)
\(\Rightarrow\frac{x}{x-1}< 1\)
\(\Rightarrow x< \frac{1}{2}\)
Tương tự: \(y< \frac{1}{2}\)
=> x+y <1
Do đó P=1
\(\frac{x}{1-x}+\frac{y}{1-y}=1\Rightarrow\frac{x.\left(1-y\right)+y\left(1-x\right)}{\left(1-x\right).\left(1-y\right)}=1\)\(\Leftrightarrow x.\left(1-y\right)+y.\left(1-x\right)=\left(1-x\right).\left(1-y\right)\)
\(\Leftrightarrow2x+2y-3xy=1\Leftrightarrow-3xy=1-2\left(x+y\right)\)(1)
ta có:\(P=x+y+\sqrt{x^2+2xy-3xy+y^2}\)
\(=x+y+\sqrt{\left(x+y\right)^2-3xy}\)(2)
Thay (1) vào (2) ta được:\(P=x+y+\sqrt{\left(x+y\right)^2-2\left(x+y\right)+1}\)
\(=x+y+\sqrt{\left(x+y-1\right)^2}=x+y-x-y+1=1\)
Vậy \(P=1\)
Bạn ơi cho tớ hỏi phần \(\sqrt{\left(x+y-1\right)^2}\) , người ta chỉ cho là x,y <1 thôi làm sao biết được x+y<1