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\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=a\)
\(\Rightarrow x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+\left(1+x^2\right)\left(1+y^2\right)=a^2\)
\(\Rightarrow x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2.x\sqrt{1+y^2}.y\sqrt{1+x^2}+1=a^2\)
\(\Rightarrow\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2+1=a^2\)
\(\Rightarrow E^2+1=a^2\)
\(\Rightarrow E=\pm\sqrt{a^2-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E^2=x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(y^2+1\right)\left(x^2+1\right)}\)
\(=2\left(xy\right)^2+x^2+y^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}\)
\(a^2=\left(xy\right)^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+\left(x^2+1\right)\left(y^2+1\right)\)
\(=2\left(xy\right)^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+x^2+y^2+1\)
\(\Rightarrow E^2=a^2-1\Rightarrow E=\sqrt{a^2-1}\)
\(E=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)
\(\Leftrightarrow E^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+y^2\right)\left(1+x^2\right)}\)
\(=2x^2y^2+x^2+y^2+2xy\left(a-xy\right)\)
\(=2x^2y^2+x^2+y^2+2xya-2x^2y^2\)
\(=x^2+y^2+2xya\)
\(=\left(2xy\right)2+a=a^2+a=E^2\)
\(E=\sqrt{a^2+a}\)
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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">
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(x^3=3+2\sqrt{2}-3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)-\left(3-2\sqrt{2}\right)\)
\(\Leftrightarrow x^3=4\sqrt{2}-3x\) \(\Leftrightarrow x^3+3x=4\sqrt{2}\) (1)
Có \(y^3=17+12\sqrt{2}-3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\right)-\left(17-12\sqrt{2}\right)\)
\(\Leftrightarrow y^3=24\sqrt{2}-3y\) \(\Leftrightarrow y^3+3y=24\sqrt{2}\) (2)
Từ (1) (2)\(\Rightarrow x^3+3x-y^3-3y=-20\sqrt{2}\)
Có \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)=\left(x-y\right)\left[\left(x-y\right)^2+3\left(xy+1\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+3\right)=x^3-y^3+3\left(x-y\right)=-20\sqrt{2}\)
Vậy \(M=-20\sqrt{2}\)
theo bài ra
\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)
\(=>x^3=\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(x^3=4\sqrt{2}-3\left[\left(\sqrt[3]{3+2\sqrt{2}}\right)\left(\sqrt[3]{3-2\sqrt{2}}\right)\right]\left[\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right]\)
\(x^3=4\sqrt{2}-3\left[\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right].x\)
\(x^3=4\sqrt{2}-3.\left[\sqrt[3]{9-\left(2\sqrt{2}\right)^2}\right]x\)
\(x^3=4\sqrt{2}-3.1x\)
\(x^3=4\sqrt{2}-3x\)
\(< =>x^3+3x-4\sqrt{2}=0\)
rồi làm y tương tự rồi thế vào M là ra
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y^2=a^2\left(1+b^2\right)+b^2\left(1+a^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(x^2=a^2b^2+\left(1+a^2\right)\left(1+b^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+1\)
\(\Rightarrow y^2+1=x^2\)
\(\Rightarrow y^2=x^2-1\)
\(\Rightarrow y=\sqrt{x^2-1}\)
\(x^2+1-y^2\ge2x\sqrt{1-y^2}\)
\(y^2+1-x^2\ge2y\sqrt{1-x^2}\)
\(\Rightarrow1\ge x\sqrt{1-y^2}+y\sqrt{1-x^2}\)
Dấu bằng xaye ra \(\Leftrightarrow\)\(x^2+y^2=1\)