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Lớn hơn hoặc bằng kí hiệu trong Latex là \geq nha!
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
\(\left(x+1\right)\left(y+1\right)=2\)
\(\Leftrightarrow x=\frac{1-y}{1+y}\)
\(P=\sqrt{x^2+y^2-\sqrt{2\left(x^2+1\right)\left(y^2+1\right)}+2}+xy\)
\(=\sqrt{\left(\frac{1-y}{1+y}\right)^2+y^2-\sqrt{2\left(\left(\frac{1-y}{1+y}\right)^2+1\right)\left(y^2+1\right)}+2}+\left(\frac{1-y}{1+y}\right)y\)
\(=\sqrt{\left(\frac{1-y}{1+y}\right)^2+y^2-2.\frac{y^2+1}{y+1}+2}+\left(\frac{1-y}{1+y}\right)y\)
\(=\sqrt{\left(\frac{y^2+1}{y+1}\right)^2}+\left(\frac{1-y}{1+y}\right)y\)
\(=\frac{y^2+1}{y+1}+\left(\frac{1-y}{1+y}\right)y=1\)
\(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2018}\)
\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)
\(x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)
\(x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2017\)
\(\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)
\(\Rightarrow A=\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)
\(\Rightarrow A=\sqrt{2017}\) khi x, y > 0 hoặc \(A=-\sqrt{2017}\) khi x, y < 0
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 \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)
\(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2009}\)
\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2009\)
\(x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2009\)
\(x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2008\)
\(\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2008\)
\(\Leftrightarrow A^2=2009\)
\(\Leftrightarrow A=\sqrt{2009}\) khi x, y > 0 hoặc \(A=-\sqrt{2009}\) khi x, y < 0
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_