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\(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2018}\)
\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)
\(x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)
\(x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2017\)
\(\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)
\(\Rightarrow A=\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)
\(\Rightarrow A=\sqrt{2017}\) khi x, y > 0 hoặc \(A=-\sqrt{2017}\) khi x, y < 0
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
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: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
nhận liên hợp ta có \(\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=x^2+1-x^2=1\)
mà theo đề bài ta có \(\left(\sqrt{x^2+1}+x\right)\left(y+\sqrt{y^2+1}\right)=1\)
==> \(\sqrt{x^2+1}-x=y+\sqrt{y^2+1}\)
tương tự ta có \(\sqrt{x^2+1}+x=\sqrt{y^2+1}-y\)
trừ từng vế 2 pt trên ta có 2x=-2y <=>x=-y
đến đây ok rùi nhé bạn
\(\text{a) }\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\\ =\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\sqrt{xy}\)
\(\text{b) }\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(\text{c) }\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}\\ =\dfrac{\sqrt{y}-1}{x-1}\)
a)\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{x}\sqrt{y}+y\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}+y\)
\(=x+\sqrt{xy}+y-x+2\sqrt{xy}+y\)
\(=3\sqrt{xy}+2y\)
Lời giải:
Ta có:
\(xy+\sqrt{(x^2+1)(y^2+1)}=2018\)
\(\Rightarrow (xy+\sqrt{(x^2+1)(y^2+1)})^2=2018^2\)
\(\Leftrightarrow x^2y^2+(x^2+1)(y^2+1)+2xy\sqrt{(x^2+1)(y^2+1)}=2018^2\)
\(\Leftrightarrow 2x^2y^2+x^2+y^2+2xy\sqrt{(x^2+1)(y^2+1)}=2018^2-1\)
\(\Leftrightarrow x^2(y^2+1)+y^2(x^2+1)+2xy\sqrt{(x^2+1)(y^2+1)}=2018^2-1\)
\(\Leftrightarrow (x\sqrt{y^2+1}+y\sqrt{x^2+1})^2=2018^2-1\)
\(\Leftrightarrow T^2=2018^2-1\)
Do đó: \(T=\pm \sqrt{2018^2-1}\)