xy2z3+x2y3z4+.......+x2014y2015z2016=? với x=-1,y=-1,z=-1
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NQ
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 5 2023
A = \(xy^2z^3+x^2y^3z^4\) + \(x^{2014}y^{2015}z^{2016}\)
Thay \(x=\) -1; y = -1; z = -1 vào A ta có:
A = (-1).(-1)2.(-1)3 + (-1)2.(-1)3.(-1)4 + (-1)2014.(-1)2015.(-1)2016
A = (-1).1(-1) + 1.(-1).1 + 1.(-1).1
A = 1 - 1 - 1
A = -1
DC
16 tháng 5 2023
A = +
Thay -1; y = -1; z = -1 vào A ta có:
A = (-1).(-1)2.(-1)3 + (-1)2.(-1)3.(-1)4 + (-1)2014.(-1)2015.(-1)2016
A = (-1).1(-1) + 1.(-1).1 + 1.(-1).1
A = 1 - 1 - 1
A = -1
tick cho mik nha
VT
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 3 2023
Với a;b;c dương ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
Lại có:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)
Áp dụng:
\(\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge\dfrac{1}{3}\left(x+y+z\right)^2.\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\)
\(=\dfrac{1}{9}\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(=\dfrac{1}{9}.9.\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dấu "=" xảy ra khi \(x=y=z\)
LT
1
XO
26 tháng 12 2020
Ta có :\(\frac{1}{x}=\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
=> \(\frac{1}{x}=\frac{y+z}{2yz}\)
=> 2yz = x(y + z)
=> 2yz - xy - xz = 0
=> (yz - xy) + (yz - xz) = 0
=> y(z - x) + z(y- x) = 0
=> y(z - x) = -z(y - x)
=> -y(x - z) = -z(y - x)
=> \(\frac{-z}{-y}=\frac{x-z}{y-x}\Leftrightarrow\frac{z}{y}=\frac{x-z}{y-x}\)
LT
1
1 tháng 6 2018
\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{1}{x+1}+1-\frac{1}{y+1}+1-\frac{1}{z+1}=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
vì \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}>=\frac{9}{x+1+y+1+z+1}=\frac{9}{1+3}=\frac{9}{4}\)(bđt svacxo)
\(\Rightarrow3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)< =3-\frac{9}{4}=\frac{3}{4}\)
dấu = xảy ra khi x=y=z=\(\frac{1}{3}\)
BD
0
TC
0
5 tháng 2 2022
\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)
\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)
\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)
A = \(xy^2z^3\) + \(x^2y^3z^4\)+...+\(x^{2014}y^{2015}z^{2016}\)
A \(\times\) \(xyz\) = \(x^2y^3z^4\)+...+\(x^{2014}y^{2015}z^{2016}\) + \(x^{2015}y^{2016}z^{2017}\)
A \(\times\) \(xyz\) - A = \(x^{2015}\)\(y^{2016}\)\(z^{2017}\) - \(xy^2z^3\)
A\(\times\)( \(xyz\) - 1) = \(x^{2015}\)\(y^{2016}z^{2017}\) - \(xy^2z^3\)
A = (\(x^{2015}\) \(y^{2016}\) \(z^{2017}\) - \(xy^2z^3\)) : (\(xyz\) - 1)
Thay \(x\) = -1; \(y\) = -1; \(z\) = -1
A = [(-1)2015.(-1)2016.(-1)2017 - (-1).(-1)2.(-1)3] : {(-1.(-1).(-1) - 1)}
A = [ 1 - 1] : [-1-1]
A = 0: (-2)
A = 0
A = ��2�3xy2z3 + �2�3�4x2y3z4+...+�2014�2015�2016x2014y2015z2016
A ×× ���xyz = �2�3�4x2y3z4+...+�2014�2015�2016x2014y2015z2016 + �2015�2016�2017x2015y2016z2017
A ×× ���xyz - A = �2015x2015�2016y2016�2017z2017 - ��2�3xy2z3
A××( ���xyz - 1) = �2015x2015�2016�2017y2016z2017 - ��2�3xy2z3
A = (�2015x2015 �2016y2016 �2017z2017 - ��2�3xy2z3) : (���xyz - 1)
Thay �x = -1; �y = -1; �z = -1
A = [(-1)2015.(-1)2016.(-1)2017 - (-1).(-1)2.(-1)3] : {(-1.(-1).(-1) - 1)}
A = [ 1 - 1] : [-1-1]
A = 0: (-2)
A = 0
Nhớ tick nha