Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(\sqrt{x^2+\sqrt{2005}}-x\right)=\sqrt{2005}\)
Kết hợp với giả thiết ta được:
\(\sqrt{x^2+\sqrt{2005}}-x=y+\sqrt{y^2+\sqrt{2005}}\)
suy ra: đpcm
2) \(\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(y+\sqrt{y^2+\sqrt{2005}}\right)=\sqrt{2005}\)
Ta có: \(\hept{\begin{cases}\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(\sqrt{x^2+\sqrt{2005}}-x\right)=\sqrt{2005}\\\left(y+\sqrt{y^2+\sqrt{2005}}\right)\left(\sqrt{y^2+\sqrt{2005}}-y\right)=\sqrt{2005}\end{cases}}\)
Kết hợp với giả thiết ta có:
\(\hept{\begin{cases}\sqrt{x^2+\sqrt{2005}}-x=y+\sqrt{y^2+\sqrt{2005}}\\\sqrt{y^2+\sqrt{2005}}-y=x+\sqrt{x^2+\sqrt{2005}}\end{cases}}\)
suy ra: \(x+y=-\left(x+y\right)\)
\(\Rightarrow\)\(S=x+y=0\)
\(a^2=b+4010\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4010\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+4010\)
\(\Rightarrow2xy+2yz+2xz=4010\Rightarrow xy+yz+xz=2005\)
\(x\sqrt{\frac{\left(2015+y^2\right)\left(2005+z^2\right)}{\left(2005+x^2\right)}}=x\sqrt{\frac{\left(xz+yz+xy+y^2\right)\left(xy+xz+yz+z^2\right)}{\left(xy+yz+x^2+xz\right)}}\)
\(=x\sqrt{\frac{\left(z\left(x+y\right)+y\left(x+y\right)\right)\left(x\left(y+z\right)+z\left(y+z\right)\right)}{\left(y\left(x+z\right)+x\left(x+z\right)\right)}}=x\sqrt{\frac{\left(y+z\right)^2\left(x+y\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}=x\left(y+z\right)=xy+xz\)
tương tự : \(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=xy+yz;z\sqrt{\frac{\left(2005+x^2\right)\left(2005+y^2\right)}{2015+z^2}}=xz+yz\)
\(\Rightarrow M=xy+xz+xy+yz+xz+yz=2\left(xy+yz+xz\right)=2\cdot2005=4010\)
\(y=\frac{1}{9+4\sqrt{5}}=\frac{1}{\left(\sqrt{5}+2\right)^2}\)
\(\Rightarrow N=\frac{1}{\left(\sqrt{5}-2\right)^2}-\frac{3}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{2}{9+4\sqrt{5}}\)
\(=\frac{1}{9-4\sqrt{5}}+\frac{2}{9+4\sqrt{5}}-3=\frac{9+4\sqrt{5}+18-8\sqrt{5}}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=24-4\sqrt{5}\)
\(S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+x^2y^2+1+x^2y^2-1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2-1\)
\(=2005^2-1\)
\(\Rightarrow S=\pm\sqrt{2005^2-1}\)
c/
Giả sử \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
\(\Leftrightarrow\sqrt[3]{3+\sqrt[3]{3}}-\sqrt[3]{3}< \sqrt[3]{3}-\sqrt[3]{3-\sqrt[3]{3}}\)
\(\Leftrightarrow\frac{\sqrt[3]{3}}{\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}}< \frac{\sqrt[3]{3}}{\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}>\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\) (1)
Ta có: \(\left\{{}\begin{matrix}\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}\\\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}>\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\end{matrix}\right.\)
Nên (1) đúng
Vậy BĐT ban đầu đúng
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_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_
22222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222
\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)(1)
Áp dụng tình chất của dấu giá trị tuyệt đối ta có:
\(\sqrt{x^2+1}>\sqrt{x^2}=\left|x\right|\ge x\)
\(\Rightarrow\sqrt{x^2+1}-x>0\)(2)
Tương tự ta có: \(\sqrt{y^2+1}-y>0\)(3)
Từ (1) và (2) \(\Rightarrow\left(x+\sqrt{x^2+1}\right)\left(\sqrt{x^2+1}-x\right)\left(y+\sqrt{y^2+1}\right)=\sqrt{x^2+1}-x\)
\(\Leftrightarrow\left(x^2+1-x^2\right)\left(y+\sqrt{y^2+1}\right)=\sqrt{x^2+1}-x\)
\(\Leftrightarrow y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\)
\(\Leftrightarrow x+y=\sqrt{x^2+1}-\sqrt{y^2+1}\)(4)
Từ (1) và (3), chứng minh tương tự ta cũng có: \(x+y=\sqrt{y^2+1}-\sqrt{x^2+1}\)(5)
Cộng vế với vế của (4) và (5) ta được:
\(2\left(x+y\right)=\left(\sqrt{x^2+1}-\sqrt{y^2+1}\right)+\left(\sqrt{y^2+1}-\sqrt{x^2+1}\right)=0\)
\(\Rightarrow x+y=0\)\(\Rightarrow x=-y\)
Thay \(x=-y\)vào A ta được: \(A=\left(-y\right)^{2005}+y^{2005}=-y^{2005}+y^{2005}=0\)