\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\Leftrightarrow\dfrac{1}{c}=\dfrac{a+b}{2ab}\)

\(\Leftrightarrow2ab=c\left(a+b\right)\)

\(\Leftrightarrow ab+ab=ca+cb\)

\(\Leftrightarrow ab-cb=ca-ab\)

\(\Leftrightarrow b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)

ab² hay là a²b²

từ a^3 + b^3 + c^3 =3abc => a+b+c = 0 

=> a+b= -c  <=> c^2 = (a+b)^2 

tương tự với -b và -a 

=> P = ab^2/a^2+b^2-a^2-2ab-b^2 + bc^2/b^2+c^2-b^2-2bc-c^2 + ca^2/c^2 + a^2 - c^2-2ac-a^2

= -a/2 - b/2 - c/2 = -1/2(a+b+c)=0

ta có :\(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{a}\)=\(\dfrac{a+b+c}{b+c+a}\)=1

*\(\dfrac{a}{b}\)=1 =>a=b

*\(\dfrac{b}{c}\)=1 =>b=c

*\(\dfrac{c}{a}\)=1 =>c=a

=>a=b=c

=>\(a^{670}\)+\(b^{672}\)+\(c^{673}\)/\(a^{2015}\)=\(a^{2015}\)/\(a^{2015}\)=1

nhớ like nha 

468 bạn nhé

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\)

Ta có: \(\frac{a^{2013}.b^2.c}{c^{2016}}=\frac{c^{2013}.c^2.c}{c^{2016}}=\frac{c^{2016}}{c^{2016}}=1\)

Câu hỏi của Hattory Heiji - Toán lớp 8 - Học toán với OnlineMath

tvbobnokb' n

iai

  ni;bv nn0

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: ...

Ta có: 
bc/a²+ac/b²+ ab/c²=abc/a³+abc/b³+abc/c³ 
=abc(1/a³ + 1/b³ + 1/c³) 
=abc[(1/a + 1/b + 1/c)(1/a² + 1/b²+ 1/c²-1/ab-1/bc-1/ca)+3/abc](áp dụng HĐt trên) 
=abc.3/(abc)=3 

tk nha bạn

thank you bạn

(^_^)