Cho a=\(\left(x+\sqrt{x^2+b}\right)\left(y+\sqrt{y^2+b}\right)\ne0\). Tính S=\(x\sqrt{y^2+b}+y\sqrt{x^2+b}\)theo a và b
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Cho a=\(\left(x+\sqrt{x^2+b}\right)\left(y+\sqrt{y^2+b}\right)\ne0\). Tính S=\(x\sqrt{y^2+b}+y\sqrt{x^2+b}\)theo a và b
Cho \(a=\left(x+\sqrt{x^2+b}\right)\left(y+\sqrt{y^2+b}\right)\ne0\) tính \(S=x\sqrt{y^2+b}+y\sqrt{x^2+b}\) theo a và b
a:dfrac{b}{left(a-4right)^2}.sqrt{dfrac{left(a-4right)^4}{b^2}}left(b0;ane4right)
b:dfrac{xsqrt{x}-ysqrt{y}}{sqrt{x}-sqrt{y}}left(xge0;yge0;xne0right)
c:dfrac{a}{left(b-2right)^2}.sqrt{dfrac{left(b-2right)^4}{a^2}left(a0;bne2right)}
d:dfrac{x}{left(y-3right)^2}.sqrt{dfrac{left(y-3right)^2}{x^2}left(x0;yne3right)}
e:2x +dfrac{sqrt{1-6x+9x^2}}{3x-1}
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a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
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a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
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Chứng minh đẳng thức:
a) dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-left(sqrt{x}-sqrt{y}right)^2sqrt{xy}left(xge0,yge0,x^2+y^2ne0right)
b) left(dfrac{1}{a-sqrt{a}}+dfrac{1}{sqrt{a}-1}right):dfrac{sqrt{a}+1}{a-2sqrt{a}+1}left(age0,ane1right)
c) sqrt{x+2sqrt{x-2}-1}left(sqrt{x-2}-1right):left(sqrt{x}-sqrt{3}right)sqrt{x}+sqrt{3}left(xge2,xne3right)
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Chứng minh đẳng thức:
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)
b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)
c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
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cho x,y,z,a là các số dương;\(a^2=b+4028và\left\{{}\begin{matrix}x+y+z=a\\x^2+y^2+z^2=b\end{matrix}\right.\).tính:
S=\(x\sqrt{\dfrac{\left(2014+y^2\right)\left(2014+z^2\right)}{2014+x^2}}\)+\(y\sqrt{\dfrac{\left(2014+z^2\right)\left(2014+x^2\right)}{2014+y^2}}\)+z\(\sqrt{\dfrac{\left(2014+x^2\right)\left(2014+y^2\right)}{2014+z^2}}\)
Ta có \(\left(x+y+z\right)^2-x^2-y^2-z^2=a^2-b\Rightarrow2\left(xy+yz+zx\right)=2048\Rightarrow xy+yz+zx=2014\)
với xy+yz+zx=2014, thay vào, ta có A=\(\sum x\sqrt{\dfrac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}=\sum x\sqrt{\dfrac{\left(y+z\right)^2\left(y+x\right)\left(z+x\right)}{\left(x+z\right)\left(x+y\right)}}=\sum x\left(y+z\right)=2\left(xy+yz+zx\right)=2048\)
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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
\(y^2=a^2\left(1+b^2\right)+b^2\left(1+a^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(x^2=a^2b^2+\left(1+a^2\right)\left(1+b^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+1\)
\(\Rightarrow y^2+1=x^2\)
\(\Rightarrow y^2=x^2-1\)
\(\Rightarrow y=\sqrt{x^2-1}\)
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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
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)
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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">
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Xem thêm câu trả lời
Rút gọn:n) N left(dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-sqrt{xy}right)left(dfrac{sqrt{x}+sqrt{y}}{x-y}right)^2o) O left(dfrac{asqrt{b}+bsqrt{a}}{sqrt{a}+sqrt{b}}+dfrac{asqrt{a}-bsqrt{b}}{sqrt{a}-sqrt{b}}right):left(dfrac{a-b}{sqrt{a}-sqrt{ }b}right)^2p) P left(dfrac{2x+1}{xsqrt{x}-1}-dfrac{sqrt{x}}{x+sqrt{x}+1}right)left(dfrac{xsqrt{x}+1}{sqrt{x}+1}-sqrt{x}right)q) Q left(dfrac{sqrt{x}+sqrt{y}}{1-sqrt{xy}}+dfrac{sqrt{x}-sqrt{y}}{1+sqrt{xy}}right):dfrac{x+xy}{1-xy}
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Rút gọn:
n) N = \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\right)^2\)
o) O = \(\left(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{a-b}{\sqrt{a}-\sqrt{ }b}\right)^2\)
p) P = \(\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)
q) Q = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\dfrac{x+xy}{1-xy}\)
n) Ta có: \(N=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\left(\dfrac{1}{\sqrt{x}-\sqrt{y}}\right)^2\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2\cdot\dfrac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
=1
o) Ta có: \(O=\left(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\right):\left(\sqrt{a}+\sqrt{b}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)
=1
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p) Ta có: \(P=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\)
\(=\sqrt{x}-1\)
q) Ta có: \(Q=\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\dfrac{x+xy}{1-xy}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{x+xy}{1-xy}\)
\(=\dfrac{x\sqrt{y}+\sqrt{x}+y\sqrt{x}+\sqrt{y}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{x+xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+xy}\)
\(=\dfrac{2\sqrt{x}\left(1+y\right)}{x\left(1+y\right)}\)
\(=\dfrac{2}{\sqrt{x}}\)
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1. tính giá trị biểu thức: B x^2-2x-frac{1-xsqrt{x}+sqrt{x}-x}{1-sqrt{x}}.frac{1+xsqrt{x}-sqrt{x}-x}{1+x} với x20172. cho 3 số dương a,b,c thỏa bne c,sqrt{a}+sqrt{b}nesqrt{c} và a+bleft(sqrt{a}+sqrt{b}-sqrt{c}right)^2.chứng minh frac{a+left(sqrt{a}-sqrt{c}right)^2}{b+left(sqrt{b}-sqrt{c}right)^2}frac{sqrt{a}-sqrt{c}}{sqrt{b}-sqrt{c}}3. cho S_kleft(sqrt{2}+1right)^k+left(sqrt{2}-1right)^kvới kin N. chứng minh S_{2009}.S_{2010}-S_{4019}2sqrt{2}4. cho x,y,z và sqrt{x}+sqrt{y}+sqrt{z}l...
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1. tính giá trị biểu thức: B = \(x^2-2x-\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}.\frac{1+x\sqrt{x}-\sqrt{x}-x}{1+x}\) với x=2017
2. cho 3 số dương a,b,c thỏa \(b\ne c,\sqrt{a}+\sqrt{b}\ne\sqrt{c}\) và \(a+b=\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)^2\).chứng minh \(\frac{a+\left(\sqrt{a}-\sqrt{c}\right)^2}{b+\left(\sqrt{b}-\sqrt{c}\right)^2}=\frac{\sqrt{a}-\sqrt{c}}{\sqrt{b}-\sqrt{c}}\)
3. cho \(S_k=\left(\sqrt{2}+1\right)^k+\left(\sqrt{2}-1\right)^k\)với \(k\in N\). chứng minh \(S_{2009}.S_{2010}-S_{4019}=2\sqrt{2}\)
4. cho x,y,z và \(\sqrt{x}+\sqrt{y}+\sqrt{z}\)là những số hữu tỉ. chứng minh \(\sqrt{x},\sqrt{y},\sqrt{z}\)là các số hữu tỉ