Cho \(a=x+y+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}b=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)Tính b theo a biết x,y>0
Những câu hỏi liên quan
Cho \(x=ab+\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\); \(y=a\sqrt{1+b^2}+b\sqrt{1+a^2}\). Tính y theo x, biết ab>0
\(y^2=a^2\left(1+b^2\right)+b^2\left(1+a^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(x^2=a^2b^2+\left(1+a^2\right)\left(1+b^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+1\)
\(\Rightarrow y^2+1=x^2\)
\(\Rightarrow y^2=x^2-1\)
\(\Rightarrow y=\sqrt{x^2-1}\)
Đúng 0
Bình luận (0)
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
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 \(x< 0,\)\(y< 0\)\(\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)
Đúng 0
Bình luận (0)
cho a=xy +\(\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)và b=x\(\sqrt{1+x^2}\)+y \(\sqrt{1+x^2}\)và x,y>0 .Tính b theo a
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 = \(\sqrt{a^2-1}\)
Đúng 0
Bình luận (0)
Rút gọn và tính giá trị biểu thức: a, frac{x+sqrt{xy}}{y+sqrt{xy}}b, frac{sqrt{a}+asqrt{b}-sqrt{b}-bsqrt{a}}{ab-1}c, frac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-left(sqrt{x}-sqrt{y}right)^2d,sqrt{frac{x-2sqrt{x}+1}{x+2sqrt{x}+1}}left(xge0right)e,frac{x-1}{sqrt{y}-1}sqrt{frac{left(y-2sqrt{y}+1right)^2}{left(x-1right)^4}}left(xne1,yne1,y0right)
Đọc tiếp
Rút gọn và tính giá trị biểu thức: a, \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
b, \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\)
c, \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
d,\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)
e,\(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(x\ne1,y\ne1,y>0\right)\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
Đúng 0
Bình luận (0)
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( 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_
Đúng 0
Bình luận (0)
\(\(d)\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)\)
\(\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\)\)
\(\(=\frac{|\sqrt{x}-1|}{|\sqrt{x}+1|}\)\)
\(\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)\)( vì \(\(x\ge0\)\))
_Minh ngụy_
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
1. Chứng minh sqrt[3]{3+sqrt[3]{3}}+sqrt[3]{3-sqrt[3]{3}} 2sqrt[3]{3}2. a) Tính Afrac{2b.sqrt{x^2-1}}{x-sqrt{x^2-1}} với xfrac{1}{2}left(sqrt{frac{a}{b}}+sqrt{frac{b}{a}}right)left(a,b0right)
b) Tính Bfrac{xy-sqrt{x^2-1}.sqrt{y^2-1}}{xy+sqrt{x^2-1}.sqrt{y^2-1}} với xfrac{1}{2}left(a+frac{1}{a}right);yfrac{1}{2}left(b+frac{1}{b}right)left(a,bge1right)3. Cho x,y thỏa mãn xyge0. Tính Bleft(left|sqrt{xy}+frac{x}{2}+frac{y}{2}right|-left|xright|right)+left(left|sqrt{xy}-frac{x}{2}-frac{y}{2}right|...
Đọc tiếp
1. Chứng minh \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
2. a) Tính \(A=\frac{2b.\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với \(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(a,b>0\right) \)
b) Tính \(B=\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) với \(x=\frac{1}{2}\left(a+\frac{1}{a}\right);y=\frac{1}{2}\left(b+\frac{1}{b}\right)\left(a,b\ge1\right)\)
3. Cho x,y thỏa mãn \(xy\ge0\). Tính \(B=\left(\left|\sqrt{xy}+\frac{x}{2}+\frac{y}{2}\right|-\left|x\right|\right)+\left(\left|\sqrt{xy}-\frac{x}{2}-\frac{y}{2}\right|-\left|y\right|\right)\)
4. Cho \(\frac{2x+2\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(x+1\right)^2}=\frac{A}{\sqrt{x}-2}+\frac{B\sqrt{x}+C}{x+1}+\frac{D\sqrt{x}+E}{\left(x+1\right)^2}\). Tìm các số A,B,C,D,E để đẳng thức trên là đúng với mọi x
Cho a= \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
b=\(x.\sqrt{1+y^2}+y.\sqrt{1+x^2}\) (x,y >0)
Tính b theo a.
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 \(x,y< 0\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)
Đúng 0
Bình luận (0)
Cho a=xy + \(\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\) và b=\(x\sqrt{1+y^2}+y\sqrt{1+x^2}\) trong đó xy>0
Tính b theo a .
Ta có:
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow b^2-a^2=-1\)
\(\Leftrightarrow b^2=a^2-1\)
Đúng 0
Bình luận (0)
12 tháng 7 2021 lúc 20:38
1. Cho \(x,y\) thỏa mãn \(\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
Tính \(x+y\)
2. Cho \(a,b\ne-2\) thỏa mãn \(\left(2a+1\right)\left(2b+1\right)=9\)
Tính \(A=\dfrac{1}{2+a}+\dfrac{1}{2+b}\)
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:
\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)
\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)
Đúng 2
Bình luận (1)
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: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)
\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)
\(=\dfrac{3+2a+1}{3a+6}\)
\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)
Đúng 1
Bình luận (2)
\(a=x\cdot y+\sqrt{\left(1+x^2\right)\cdot\left(1+y^2\right)}\) \(b=x\cdot\sqrt{1+y^2}+y\cdot\sqrt{1+x^2}\) với xy>0 tính b theo a
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow a^2-b^2=1\)
\(\Rightarrow a^2=1+b^2\)
Đúng 0
Bình luận (0)