Cho x,y ∈R , \(\left(x+\sqrt{a+x^2}\right)\left(y+\sqrt{a+y^2}\right)\)
Tính K= \(\left(x+\sqrt{a+y^2}\right)\left(y+\sqrt{a+x^2}\right)\)
Những câu hỏi liên quan
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)
Đọc tiếp
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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Khai triển và rút gọn biểu thức ( x ≥ 0, y ≥ 0 )
a, \(\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
b, \(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{x}\sqrt{y}+y\right)\)
c, \(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)
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|...
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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
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}
Đọc tiếp
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: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
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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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Khai triển và rút gọn biểu thức (với x và y > 0):
a)\(\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
b)\(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)
c)\(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)
Giải:
a) \(\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}\right)^3-1\)
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b) \(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)
\(=\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3\)
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c) \(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)
\(=6x+3\sqrt{xy}-4\sqrt{xy}-2y\)
\(=6x-\sqrt{xy}-2y\)
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\(a.\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)=x\sqrt{x}-1\)
\(b.\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)=x\sqrt{x}+y\sqrt{y}\)
\(c.\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)=6x-\sqrt{xy}-2y\)
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Biết \(0< x\le y\)và \(\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)+2\left(x+2y\right)}\right)+\left(\frac{y}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}\right)=\frac{5}{3}\)
Tính \(\frac{x}{y}\)
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:
\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)
\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)
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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: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)
\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)
\(=\dfrac{3+2a+1}{3a+6}\)
\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)
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giải hệ pt :
a,\(\left\{{}\begin{matrix}\sqrt{y}\left(\sqrt{x}+\sqrt{x+3}\right)=3\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x^2+x=y^2+y\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^4+y^4+x^2y^2=21\end{matrix}\right.\)
a, ĐK: \(x,y\ge0\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\sqrt{y}}{\sqrt{x+3}-\sqrt{x}}=3\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=\sqrt{x+3}\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+3}=x+1\)
\(\Leftrightarrow x+3=x^2+2x+1\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\left(l\right)\end{matrix}\right.\)
Thay \(x=1\) vào hệ phương trình đã cho ta được \(y=1\)
Vậy pt đã cho có nghiệm \(x=y=1\)
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b, \(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(y+\dfrac{1}{2}\right)^2\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y=-1\end{matrix}\right.\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-3x=0\end{matrix}\right.\left(1\right)\\\left\{{}\begin{matrix}x+y=-1\\x^2+y^2=-3\end{matrix}\right.\left(vn\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=0\end{matrix}\right.\)
Vậy ...
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c, Đặt \(\left\{{}\begin{matrix}x^2+y^2=a\\xy=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\a^2-b^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\a-b=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2=9\)
\(\Rightarrow x+y=\pm3\)
TH1: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=-3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
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Cho x, y, z > 0 và xy + yz + zx = a.
Chứng minh: \(x\sqrt{\frac{\left(a+y^2\right)\left(a+z^2\right)}{a+x^2}}+y\sqrt{\frac{\left(a+z^2\right)\left(a+x^2\right)}{a+y^2}}+z\sqrt{\frac{\left(a+x^2\right)\left(a+y^2\right)}{a+z^2}}=2a\)
Lời giải:
\(xy+yz+xz=a\Rightarrow \left\{\begin{matrix} x^2+a=x^2+xy+yz+xz=(x+y)(x+z)\\ y^2+a=y^2+xy+yz+xz=(y+x)(y+z)\\ z^2+a=z^2+xy+yz+xz=(z+x)(z+y)\end{matrix}\right.\)
Khi đó:
\(x\sqrt{\frac{(a+y^2)(a+z^2)}{a+x^2}}+y\sqrt{\frac{(a+z^2)(a+x^2)}{a+y^2}}+z\sqrt{\frac{(a+x^2)(a+y^2)}{a+z^2}}\)
\(=x\sqrt{\frac{(y+x)(y+z)(z+x)(z+y)}{(x+y)(x+z)}}+y\sqrt{\frac{(z+x)(z+y)(x+y)(x+z)}{(y+x)(y+z)}}+z\sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}\)
\(=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2a\)
Ta có đpcm.
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