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\(y^2=a^2\left(1+b^2\right)+b^2\left(1+a^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(x^2=a^2b^2+\left(1+a^2\right)\left(1+b^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+1\)
\(\Rightarrow y^2+1=x^2\)
\(\Rightarrow y^2=x^2-1\)
\(\Rightarrow y=\sqrt{x^2-1}\)
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 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_
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:
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow b^2-a^2=-1\)
\(\Leftrightarrow b^2=a^2-1\)
Bài 1.
Ta có:\(\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)
\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)\)
\(\Rightarrow y+\sqrt{y^2+2020}=\sqrt{x^2+2020}-x\)
\(\Rightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\) (1)
Ta có:\(\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)
\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)\)
\(\Rightarrow x+\sqrt{x^2+2020}=\sqrt{y^2+2020}-y\)
\(\Rightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\) (2)
Cộng vế với vế của (1) và (2) ta có:
\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)
\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)
Bài 2:
Ta có: (2a+1)(2b+1)=9
nên \(2b+1=\dfrac{9}{2a+1}\)
\(\Leftrightarrow2b=\dfrac{9}{2a+1}-\dfrac{2a+1}{2a+1}=\dfrac{8-2a}{2a+1}\)
\(\Leftrightarrow b=\dfrac{8-2a}{4a+2}=\dfrac{4-a}{2a+1}\)
\(\Leftrightarrow b+2=\dfrac{4-a+4a+2}{2a+1}=\dfrac{3a+6}{2a+1}\)
Ta có: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)
\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)
\(=\dfrac{3+2a+1}{3a+6}\)
\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow a^2-b^2=1\)
\(\Rightarrow a^2=1+b^2\)
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
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