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câu 2 :
ab+ bc + ca = 2015
=> 2015 +a^2 = a^2 + ab + bc + ca
=> 2015 + a^2 = a(a+b ) + c( a + b ) = ( a + c )( a + b)
Tương tự : 2015+b^2 = ( b + c )(b +a )
2015 + c^2 = ( c + a )(c + b ) thay vào ta có :
( 2015 + a^2)(2015 + b^2 ) (2015 +c^2) = (a + c )(a+b)(b+c)(b+a)(c+a)(c+b) = [(a+c)(a+b)(b+c) ]^2 là số chính phương
Câu 1 ) :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2015}\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{2015}-\frac{1}{z}=\frac{z-2015}{2015z}\)
=> \(\frac{x+y}{xy}=\frac{z-2015}{2015z}\)
=> \(2015z\left(x+y\right)=\left(z-2015\right)xy\)
=> \(2015z\left(2015-z\right)-\left(z-2015\right)xy\) = 0
=> \(\left(2015-z\right)\left(2015z-xy\right)\)= 0
=> \(\left(2015-z\right)\left(2015\left(2015-x-y\right)-xy\right)=0\)
=> \(\left(2015-z\right)\left(2015^2-2015x-2015y-xy\right)=0\)
=> \(\left(2015-z\right)\left(2015-x\right)\left(2015-y\right)=0\)
=> 2015 - z = 0 hoặc 2015 -x = 0 hoặc 2015 - y = 0
=> z = 2015 hoặc x= 2015 hoặc y = 2015
Vậy trong ba số có ít nhất 1 số bằng 2015
Từ giả thiết ta có ngay \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
Suy ra x + y = 0 hoặc y + z = 0 hoặc z + x = 0
Tới đây bạn tự làm nhé :)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\left(x;y;z,x+y+z\ne0\right)\)
\(\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)
\(\Rightarrow\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)
\(\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\)
\(\Leftrightarrow\left(xy+yz\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow y\left(x+z\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left(x+y\right)\left(y+z\right)=0\)
Từ đó \(x=-z\)hoặc \(x=-y\)hoặc \(y=-z\)
-Nếu \(x=-z\Rightarrow z^{2017}+x^{2017}=0\Rightarrow M=\frac{19}{4}+0=\frac{19}{4}\)
Tương tự với các trường hợp còn lại, ta cũng tính được \(M=\frac{19}{4}\)
Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành:
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)
\(\Rightarrow x=2018;y=2019;z=2020\)
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)
\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)
\(x=2018,y=2019,z=2020\)
Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)
\(\sqrt{y-2014}=b\left(b>0\right)\)
\(\sqrt{z-2015}=c\left(c>0\right)\)
Có \(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)
<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)
<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)
<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).
Có \(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)
\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)
\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)
=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)
Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)
Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xyz}\left(x+y+z\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)(vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\))
Mặt khác, ta có : \(\frac{1}{x+y+z}=2\) .
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
=> x+y = 0 hoặc y + z = 0 hoặc z + x = 0
Từ đó suy ra P = 0 (lí do vì x,y,z là các số mũ lẻ)
Áp dụng BĐT Cauchy-Schwarz dạng Engelta có:
\(VT=\frac{700}{2\left(xy+yz+xz\right)}+\frac{386}{x^2+y^2+z^2}\)\(=\frac{\sqrt{700}^2}{2\left(xy+yz+xz\right)}+\frac{\sqrt{386}^2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{700}+\sqrt{386}\right)^2}{x^2+y^2+z^2+2\left(xy+yz+xz\right)}\)\(=\frac{\left(\sqrt{700}+\sqrt{386}\right)^2}{\left(x+y+z\right)^2}\)
\(=\left(\sqrt{700}+\sqrt{386}\right)^2>2015\left(x+y+z=1\right)\)
Ta có\(x\sqrt{\frac{\left(2015+y^2\right)\left(2015+z^2\right)}{2015+x^2}}=x\sqrt{\frac{\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}{xy+yz+zx+x^2}}\)
\(=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=xy+xz\)
Tương tự:\(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=yx+yz\)
\(z\sqrt{\frac{\left(2015+x^2\right)\left(2015+y^2\right)}{2015+z^2}}=zx+zy\)
Ta có :\(P=xy+xz+yx+yz+zx+zy=2\left(xy+yz+zx\right)=4030\)
=>P không phải là số chính phương
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2015\Leftrightarrow\frac{1}{x}+\frac{1}{y}=2015-\frac{1}{z}=\frac{z-2015}{2015z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-2015}{2015z}\Leftrightarrow2015z\left(x+y\right)=xy\left(z-2015\right)\)
\(2015z\left(2015-z\right)+\left(2015-z\right)xy=0\Leftrightarrow\left(2015-z\right)\left(2015z+xy\right)=0\)
\(\Leftrightarrow\left(2015-z\right)\left(2015\left(2015-x-y\right)+xy\right)=0\)
\(\Leftrightarrow\left(2015-z\right)\left(2015^2-2015x-2015y+xy\right)=0\)
\(\Leftrightarrow\left(2015-z\right)\left(2015-y\right)\left(2015-x\right)=0\)
vậy trong 3 số sẽ có 1 số là 2015
nguyen thi thu thuy copy ac nhi?con doi tick nua chu!!!du sao cung thong minh nen tuj tick cho :V