Cho a,b,x,y thuộc z mà.
a+b=x+y.
a^2+b^2=x^2+y^2.
CMR a^2003+b^2003=x^2003+y^2003
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TT
0
18 tháng 8 2017
\(\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
=\(\left(\dfrac{x^2}{a^2}-\dfrac{x^2}{a^2+b^2+c^2}\right)+\left(\dfrac{y^2}{b^2}-\dfrac{y^2}{a^2+b^2+c^2}\right)\)+\(\left(\dfrac{z^2}{c^2}-\dfrac{z^2}{a^2+b^2+c^2}\right)=0\)
=\(x^2.\dfrac{b^2+c^2}{a^2+b^2+c^2}+y^2.\dfrac{a^2+c^2}{a^2+b^2+c^2}+z^2.\dfrac{a^2+b^2}{a^2+b^2+c^2}=0\)
Vì \(a,b,c\) \(\ne\)0 nên dấu "=" xảy ra khi \(x=y=z=0\)
\( \Rightarrow\)\(A=x^{2003}+y^{2003}+z^{2003}=0+0+0=0\)
Chúc Bạn Học Tốt !!!
S
14 tháng 2 2020
\(x+\left(x+1\right)+....+2003=2003\Leftrightarrow x+\left(x+1\right)+....+2002=0\)
\(\Leftrightarrow\left(2002+x\right)\left(2002-x+1\right)=0\Leftrightarrow\left(2002+x\right)\left(2003-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2002\\x=2003\end{cases}}\)
TM
0
TH
28 tháng 8 2018
a) Ta có:
\(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
\(=-5n\)
Vì \(-5n⋮5\) với n thuộc Z
\(\Rightarrow n\left(2n-3\right)-2n\left(n+1\right)⋮5\) với n thuộc Z
b) Ta có:
\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)
\(=5n^2+5n\)
\(=5\left(n^2+n\right)\)
Vì \(5\left(n^2+n\right)⋮5\)
\(\Rightarrow\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)
c) Ta có:
\(\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1-2\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1\right)\left(x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1\right)\left(x^{2003}+y^{2003}-x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)\)
\(=2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)\)
Vì \(2\left(xy+1\right)y^{2003}⋮2\)
\(2\left(x^{2003}+y^{2003}\right)⋮2\)
\(\Rightarrow2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)⋮2\)
\(\Rightarrow\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)⋮2\)
22 tháng 8 2022
a: \(=n^3+2n^2+3n^2+6n-n-2-n^3+5\)
\(=5n^2+5n+3⋮̸5\)
b:\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
d: \(=4x^2y^2-2x^2y+2xy^2-xy-4x^2y^2+xy\)
\(=-2\left(x^2y-xy^2\right)⋮2\)
VM
1 tháng 11 2019
a) \(\left|x+\frac{13}{17}\right|+\left|y+\frac{2019}{2018}\right|+\left|z-2007\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\frac{13}{17}\right|\ge0\\\left|y+\frac{2019}{2018}\right|\ge0\\\left|z-2007\right|\ge0\end{matrix}\right.\forall x,y,z.\)
\(\Rightarrow\left|x+\frac{13}{17}\right|+\left|y+\frac{2019}{2018}\right|+\left|z-2007\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\frac{13}{17}=0\\y+\frac{2019}{2018}=0\\z-2007=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-\frac{13}{17}\\y=0-\frac{2019}{2018}\\z=0+2007\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{13}{17}\\y=-\frac{2019}{2018}\\z=2007\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\in\left\{-\frac{13}{17};-\frac{2019}{2018};2007\right\}.\)
Chúc bạn học tốt!
\(a+b=x+y\Leftrightarrow a-x=y-b\)
\(a^2+b^2=x^2+y^2\Leftrightarrow\left(a-x\right)\left(a+x\right)=\left(y-b\right)\left(y+b\right)\)
mà a-x = y-b\(\Rightarrow a+x=b+y\)
lại có a+b =x+y => 2a+b+x=2y+b+x=> a=y
suy ra b=x