Chứng minh : a = xy + \(\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\) , b=\(x\sqrt{1+y^2}+y\sqrt{1+x^2}\)
Tính b theo a nếu x,y>0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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
30 tháng 7 2022
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
3 tháng 7 2019
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
3 tháng 7 2019
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( 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 a2 = 2x2 y2 + x2 + y2 + 1 + \(2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
b2 = 2x2 y2 + x2 + y2 + \(2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
Từ đó => a2 = b2 + 1
=> b = \(\sqrt{a^2-1}\)
D
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 \(x< 0,\)\(y< 0\)\(\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)
9 tháng 7 2022
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: \(1< \left(1+x^2\right)\left(1+y^2\right)\)\(\Leftrightarrow\)\(1< xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)\(\Leftrightarrow\)\(a^2>1\)