Cho \(A=xy.\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(B=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)
Điều kiện: xy >0
Tính b theo a
K
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8 tháng 6 2018
Ta có:
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow b^2-a^2=-1\)
\(\Leftrightarrow b^2=a^2-1\)
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: \(1< \left(1+x^2\right)\left(1+y^2\right)\)\(\Leftrightarrow\)\(1< xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)\(\Leftrightarrow\)\(a^2>1\)
25 tháng 9 2018
Ta co: \(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=y+z\)
Thê vào ta được
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
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 = \(\sqrt{a^2-1}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2019
\(b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+x^2y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2-1\)
\(=a^2-1\Rightarrow b=\sqrt{a^2-1}\)
21 tháng 10 2016
Bài 1
a, \(\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
=\(\left(\sqrt{y}+\sqrt{x}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
b,\(\sqrt{8+2.2\sqrt{2}+1}-\sqrt{8-2.2\sqrt{2}+1}\)
=\(\sqrt{\left(\sqrt{8}+1\right)^2}-\sqrt{\left(\sqrt{8}-1\right)^2}\)
=\(\sqrt{8}+1-\left(\sqrt{8}-1\right)\)
=2
Bài 2
a, ĐKXĐ : x\(\ge\)0, x\(\pm\)1
b, Q=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}+x+\sqrt{x}-x}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{3-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-3+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{3\sqrt{x}-3}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{-3}{1+\sqrt{x}}\)
c, de Q = 2 => \(\frac{-3}{1+\sqrt{x}}\)=2 =>1+\(\sqrt{x}\)=-6 =>\(\sqrt{x}\)=-7 =>x vô nghiệm
21 tháng 10 2016
Bài 1: \(\left(\frac{\sqrt{xy}-\sqrt{y}}{\sqrt{x}-1}+\frac{\sqrt{xy}-\sqrt{x}}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}-\sqrt{y}\right)\)
\(\sqrt{9+4\sqrt{2}}-\sqrt{9-4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =2\sqrt{2}+1-2\sqrt{2}+1=2\)
21 tháng 10 2016
Bài 2:
\(Q=\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{-\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-x-\sqrt{x}+x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3}{\sqrt{x}+1}\)
Để Q=2
=> \(\frac{-3}{\sqrt{x}+1}=2\)
\(\Leftrightarrow2\left(\sqrt{x}+1\right)=-3\)
\(\Leftrightarrow2\sqrt{x}+2=-3\)
\(\Leftrightarrow2\sqrt{x}=-5\) (vô lí)
Vậy k có giá trị nào của x thỏa mãn Q=2
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 \(x< 0,\)\(y< 0\)\(\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)
Bạn xem lại đề bài xem \(A=xy\sqrt{(x^2+1)(y^2+1)}\) hay \(A=xy+\sqrt{(x^2+1)(y^2+1)}\) thế?