Thử nhé

Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)

Thay vo P ta duoc \(P=4.\sqrt{2021}\)

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

Cauchy-Schwarz:

\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)

\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)

\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)

Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)

\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)

\(B=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{xy}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\left(\sqrt{x^3}+x\right)}=\dfrac{\left(\sqrt{x}+1\right)\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)x\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}}\)

Đặt \(\sqrt{x}=a\) , \(\sqrt{y}=b\) , \(\sqrt{z}=c\)

Suy ra \(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Xét tử : \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left[-\left(a-b\right)-\left(c-a\right)\right]+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(c^2-a^2\right)+\left(c-a\right)\left(b^2-a^2\right)=\left(a-b\right)\left(c-a\right)\left(c+a\right)+\left(c-a\right)\left(b-a\right)\left(b+a\right)\)

\(=\left(a-b\right)\left(c-a\right)\left(c+a-a-b\right)=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

Suy ra \(P=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)

\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)

\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)

\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)

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

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)

\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)

\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé 

\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)

\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)

 Linh ơi, câu a,b,c bạn làm đều đúng hết kết quả cách làm đều đúng nhưng mà ở chỗ câu c): \(\sqrt{x}^3+\sqrt{y}^3\)

không phải vậy đâu, mặc dù mình biết bạn hiểu, hay do sơ suất, nhưng mà chỗ đó là \(\sqrt{x^3}+\sqrt{y^3}\)nha! Dù sao cũng cảm ơn bạn nha!

\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\)\(\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)

\(=-\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{z}-\sqrt{x}\right)}-\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(-\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\frac{-x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{z}-\sqrt{x}\right)-z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-x\sqrt{y}+x\sqrt{z}-y\sqrt{z}+y\sqrt{x}-z\sqrt{x}+z\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{z}\left(x-y\right)-z\left(\sqrt{x}-y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-\sqrt{xy}+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-\sqrt{xy}+\sqrt{xz}+\sqrt{yz}-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{\sqrt{y}\left(\sqrt{z}-\sqrt{x}\right)-\sqrt{z}\left(\sqrt{z}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

Với x, y > 0 có:

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