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B> \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\)
\(\Leftrightarrow\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(\left(x-\sqrt{x^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow y+\sqrt{y^2+2013}=-x+\sqrt{x^2+2013}\)
Chứng minh tương tự: \(x+\sqrt{x^2+2013}=-y+\sqrt{y^2+2013}\)
cộng vế theo vế ta được: \(x+y=-x-y\)
\(\Leftrightarrow x+y=0\Leftrightarrow x=-y\Leftrightarrow x^{2013}=-y^{2013}\)
\(\Leftrightarrow x^{2013}+y^{2013}=0\)
a,Ta có x =...
x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3+1}-1}\right)}{\left(\sqrt{\sqrt{3}+1}\right)\left(\sqrt{\sqrt{3}-1}\right)}\)
x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}\)
x = \(\frac{\sqrt{3}.2}{\sqrt{3}}\)
x = 2
sau đó thay x=2 vào A nhé.
A=2014 !!!
Ta có
\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)
Thế vào ta được
\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)
\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)
Bạn thêm điều kiện x,y,z lớn hơn 0 nhé :)
Từ giả thiết ta suy ra : \(a^2=b+4032\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4032\)
\(\Rightarrow xy+yz+zx=2016\)thay vào :
\(x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}\)
\(=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+y\right)\left(z+x\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=x\left|y+z\right|=xy+xz\)vì x,y,z > 0
Tương tự : \(y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{2016+y^2}}=xy+zy\)
\(z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{2016+z^2}}=zx+zy\)
Suy ra \(P=2\left(xy+yz+zx\right)=2.2016=4032\)
\(\left(x+\sqrt{x^2+\sqrt{2013}}\right)\left(x-\sqrt{x^2+\sqrt{2013}}\right)=x^2-x^2-\sqrt{2013}=-\sqrt{2013}\) (1)
Theo đề bài và (1) => dpcm
b) theo a có \(y+\sqrt{y^2+\sqrt{2013}}=-x+\sqrt{x^2+\sqrt{2013}}\)(2)
tương tự ta có \(x+\sqrt{x^2+\sqrt{2013}}=-y+\sqrt{y^2+\sqrt{2013}}\)(3)
Cộng 2 vế (2) với (3) => x+y = -x -y
hay 2(x+y) =0 =>S= x+y =0
b) đk: \(x>2012;y>2013\)
pt \(\frac{16}{\sqrt{x-2012}}+\sqrt{x-2012}+\frac{1}{\sqrt{y-2013}}+\sqrt{y-2013}=10\)
\(VT\ge2\sqrt{\frac{16}{\sqrt{x-2012}}.\sqrt{x-2012}}+2\sqrt{\frac{1}{\sqrt{y-2013}}.\sqrt{y-2013}}=8+2=10\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-2012=16\\y-2013=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2028\\y=2014\end{cases}}\)
Sửa đề: chứng minh
\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=4-a\)
ĐKXĐ: \(a\ge0;a\ne1\)
\(\left(\frac{2\left(\sqrt{a}-1\right)+a-\sqrt{a}}{\sqrt{a}-1}\right)\left(\frac{2\left(\sqrt{a}+1\right)-a-\sqrt{a}}{\sqrt{a}+1}\right)\)
\(=\left(\frac{\sqrt{a}+a-2}{\sqrt{a}-1}\right)\left(\frac{\sqrt{a}-a+2}{\sqrt{a}+1}\right)\)
\(=\frac{\left(\sqrt{a}+a-2\right)\left(\sqrt{a}-a+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{\left(\sqrt{a}\right)^2-\left(a-2\right)^2}{\left(\sqrt{a}\right)^2-1}\)
\(=\frac{a-a^2+4a-4}{a-1}\)
\(=\frac{-a\left(a-1\right)+4\left(a-1\right)}{a-1}\)
\(=\frac{\left(4-a\right)\left(a-1\right)}{a-1}=4-a=VP\)
=> đpcm
\(\sqrt{a+2016}\)- \(\sqrt{a+2013}\) = (\(\sqrt{a+2016}\)- \(\sqrt{a+2013}\)) . (\(\sqrt{a+2016}+\)\(\sqrt{a+2013}\)) / \(\sqrt{a+2016}+\)\(\sqrt{a+2013}\)[ nhân cả tử và mẫu với (\(\sqrt{a+2016}+\)\(\sqrt{a+2013}\)), (mẫu cũ =1) ]
= (a+2016)-(a+2013)/\(\sqrt{a+2016}+\)\(\sqrt{a+2013}\)
k mk nha
\(\frac{\left(a+2016\right)-\left(a+2013\right)}{\sqrt{a+2016}+\sqrt{a+2013}}\)\(=\frac{\left(\sqrt{a+2016}\right)^2-\left(\sqrt{a+2013}\right)^2}{\sqrt{a+2016}+\sqrt{a+2013}}=\frac{\left(\sqrt{a+2016}+\sqrt{a+2013}\right)\left(\sqrt{a+2016}-\sqrt{a-2013}\right)}{\sqrt{a+2016}+\sqrt{a+2013}}\)
\(=\sqrt{a+2016}-\sqrt{a+2013}\)