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2 tháng 10 2017

1.

\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)

hết tối giải rồi

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

b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_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_

8 tháng 6 2018

Ta có:

\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)

\(\Rightarrow b^2-a^2=-1\)

\(\Leftrightarrow b^2=a^2-1\)

4 tháng 9 2016

Ta có a= 2x2 y+ x2 + y2 + 1 + \(2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

b= 2x2 y2 + x2 + y+ \(2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

Từ đó => a= b+ 1

=> b = \(\sqrt{a^2-1}\)

1 tháng 12 2017

Xét \(a^2=x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+\left(1+x^2\right)\left(1+y^2\right)\)

\(b^2=x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+y^2\left(1+x^2\right)\)

\(\Rightarrow b^2=a^2-1\)

Nếu \(x>0,y>0\Rightarrow b>0\Rightarrow b=\sqrt{a^2-1}\)

Nếu \(x< 0,\)\(y< 0\)\(\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)

Bài 1: 

a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

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

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

b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)

Do đó: A>=0

1 tháng 9 2019

\(b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

\(=x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

Mà \(a^2=x^2+y^2+2x^2y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

\(\Leftrightarrow\)\(2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=a^2-\left(x^2+y^2+2x^2y^2\right)-1\)

\(\Rightarrow\)\(b^2=x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

\(=x^2+y^2+2x^2y^2+a^2-\left(x^2+y^2+2x^2y^2\right)-1=a^2-1\)\(\Leftrightarrow\)\(b=\sqrt{a^2-1}\) ( do a2>1 ) 

Cm: \(a^2>1\)

Có: \(1< \left(1+x^2\right)\left(1+y^2\right)\)\(\Leftrightarrow\)\(1< xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)\(\Leftrightarrow\)\(a^2>1\)