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a) +) Điều kiện : x \(\ge\) 0 ; y \(\ge\) 0 ; y \(\ne\) 1 ; x; y không đồng thời bằng 0
+) \(P=\frac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{x\sqrt{x}+x-y+y\sqrt{y}-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+y-\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(x+\sqrt{x}\right)+\left(y-xy\right)-\left(\sqrt{xy}+\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(1+\sqrt{x}\right)\sqrt{x}+y\left(1-x\right)-\sqrt{y}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(1+\sqrt{x}\right)\left(\sqrt{x}+y-y\sqrt{x}-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-y\sqrt{x}\right)+\left(y-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}=\frac{\sqrt{x}\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}\)
\(P=\sqrt{x}\left(1+\sqrt{y}\right)-\sqrt{y}=\sqrt{x}-\sqrt{y}+\sqrt{xy}\)
b) Để P = 2 <=> \(\sqrt{x}-\sqrt{y}+\sqrt{xy}=2\) <=> \(\sqrt{x}+\sqrt{xy}=\sqrt{y}+2\)
<=> \(\left(\sqrt{x}+\sqrt{xy}\right)^2=\left(\sqrt{y}+2\right)^2\)
<=> \(x+xy+2x\sqrt{y}=y+4+4\sqrt{y}\)
<=> \(x+xy-y+\left(2x-4\right)\sqrt{y}=4\)(*)
P = 2 <=> (x; y) thỏa mãn (*)
Ta có : \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)
\(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)
nhân theo vế của ( 1 ) ; ( 2 ) , ta có :
\(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)
\(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)
rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :
\(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)
\(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\)
\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
A = 2017
( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :) )
2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)
\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)
\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)
Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)
Thay \(1=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) ta có
\(1+x=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Tương tự \(1+y=\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\) và \(1+z=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)\)
\(\Rightarrow\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
và \(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{z}+\sqrt{x}\right)+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
Do đó P = 2
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Tên youtube:P Music
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Vanh_GoG_VN
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