cho a, b, c nguyên thỏa mãn: a3+b3+c3=3abc
tính \(S=\dfrac{a^{2017}}{b^{2017}}+\dfrac{b^{2017}}{c^{2017}}+\dfrac{c^{2017}}{a^{2017}}\)
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T
20 tháng 11 2023
Có:
\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)
Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)
Thay \(a=b=c\) vào \(A\), ta được:
\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)
\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)
\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)
\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)
\(=\dfrac{3}{2017^2}\)
Vậy: ...
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2021
\(1-\dfrac{1}{1+a}\ge\dfrac{2017}{b+2017}+\dfrac{2018}{c+2018}\ge2\sqrt{\dfrac{2017.2018}{\left(b+2017\right)\left(c+2018\right)}}\)
\(1-\dfrac{2017}{b+2017}\ge\dfrac{1}{1+a}+\dfrac{2018}{b+2018}\ge2\sqrt{\dfrac{2018}{\left(1+a\right)\left(b+2018\right)}}\)
\(1-\dfrac{2018}{c+2018}\ge\dfrac{1}{1+a}+\dfrac{2017}{b+2017}\ge2\sqrt{\dfrac{2017}{\left(1+a\right)\left(b+2017\right)}}\)
Nhân vế:
\(\dfrac{abc}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\ge\dfrac{8.2017.2018}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\)
\(\Rightarrow abc\ge8.2017.2018\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2021
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2.1;2.2017;2.2018\right)=...\)
2 tháng 6 2022
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{b^{2017}\cdot k^{2017}+d^{2017}\cdot k^{2017}}{b^{2017}+d^{2017}}=k^{2017}\)
\(\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}=\dfrac{\left(bk+dk\right)^{2017}}{\left(b+d\right)^{2017}}=k^{2017}\)
Do đó: \(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
FT
11 tháng 10 2017
Đặt:\(\dfrac{a}{b}=\dfrac{c}{d}=@\Leftrightarrow\left\{{}\begin{matrix}a=b@\\c=d@\end{matrix}\right.\)
khi đó: \(\dfrac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\dfrac{b^{2017}@^{2017}+b^{2017}}{d^{2017}@^{2017}+d^{2017}}=\dfrac{b^{2017}\left(@^{2017}+1\right)}{d^{2017}\left(@^{2017}+1\right)}=\dfrac{b^{2017}}{d^{2017}}\)
\(\dfrac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\dfrac{\left(b@-b\right)^{2017}}{\left(d@-d\right)^{2017}}=\dfrac{\left[b\left(@-1\right)\right]^{2017}}{\left[d\left(@-1\right)\right]^{2017}}=\dfrac{b^{2017}}{d^{2017}}\)
Ta có điều phải chứng minh
3 tháng 1 2017
\(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\frac{\left(a+b+c\right)}{2}.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow a=b=c\) (a,b,c là các số dương)
Bạn thay vào A để tính.
17 tháng 8 2017
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a+c+a-c}{b+d+b-d}=\dfrac{2a}{2b}=\dfrac{a}{b}\left(1\right)\)
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a+c-a+c}{b+d-b+d}=\dfrac{2c}{2d}=\dfrac{c}{d}\left(1\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Thay vào tính
máu biếng tới tận não:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\left[\left(a+b\right)^3+c^2\right]-ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\dfrac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{2}=0\)
\(\Leftrightarrow\left(a+b+c\right)\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a-b=b-c=c-a\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Mà a,b,c >0
=> a = b = c
=> S = 3
\(\)
sao mấy bn giỏi wá z
Mấy bài này mik chả hỉu j cả T^T