Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 = \(\sqrt{a^2-1}\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_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_
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 \(x< 0,\)\(y< 0\)\(\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)
Ta có:
\(a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Rightarrow a^2=x^2+y^2+x^2y^2+1\)
\(b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Rightarrow b^2=x^2+y^2+x^2y^2\)
\(\Rightarrow b^2=a^2-1\)
Nếu \(x,y>0\Rightarrow b>0\Rightarrow b=\sqrt{a^2-1}\)
Nếu \(x,y< 0\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)
Ta có:
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow b^2-a^2=-1\)
\(\Leftrightarrow b^2=a^2-1\)
\(b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+x^2y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2-1\)
\(=a^2-1\Rightarrow b=\sqrt{a^2-1}\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow a^2-b^2=1\)
\(\Rightarrow a^2=1+b^2\)
Đề đúng : Cho \(a=xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\) , \(b=x\sqrt{1+y^2}+y\sqrt{1+x^2}\). Hãy tính b theo a, biết x,y> 0
Giải :
Ta có : \(a^2=\left(xy\right)^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+2x^2y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=a^2-1\)
Vậy \(b=\sqrt{a^2-1}\)(vì x,y> 0 nên b > 0)
khó quá đi em mới học lớp 6 thôi hu hu
<img class="irc_mi i5I_Ps3Xg92k-pQOPx8XEepE" alt="" style="margin-top: 100px;" src="http://dungfacebook.net/wp-content/uploads/2015/11/622.jpg" width="304" height="196">