a) \(a^2+b^2+c^2=ab+bc+ac\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)

Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(c-a\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)

\(\Leftrightarrow a=b=c\)

a. \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ab-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)

c) \(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}+\frac{x+2}{2018}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}+\frac{x+2018}{2}\)

Ta có VT + 4 = VP + 4

VT + 4 = \(\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+2}{2018}+1\right)\) 

\(=\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}+\frac{x+2020}{2018}\)

\(=\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)

VP + 4 = \(\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)+\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}\right)\)

\(=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}+\frac{x+2020}{2}\)

\(=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)

Khi đó \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)

=> \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

Vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)

=> x + 2020 = 0

=> x = -2020

\(a^2+b^2+c^2=\frac{b^2-c^2}{a^2+3}+\frac{c^2-a^2}{b^2+4}+\frac{a^2-b^2}{c^2+5}\)

<=>\(a^2-\frac{a^2-b^2}{c^2+5}+b^2-\frac{b^2-c^2}{a^2+3}+c^2-\frac{c^2-a^2}{b^2+4}=0\)

<=>\(\frac{ac^2+4a^2+b^2}{c^2+5}+\frac{ba^2+4b^2+c^2}{a^2+3}+\frac{ab^2+4c^2+a^2}{b^2+4}=0\)

Vì \(VT\ge0\) nên dấu "=" xảy ra khi a=b=c=0 => S = 2017 + bc + 20c=2017+0.0+20.0=2017

Ta có \(\frac{2a+b+c}{b+c}=\frac{2b+c+a}{c+a}=\frac{2c+a+b}{a+b}\Rightarrow\frac{2a}{b+c}+1=\frac{2b}{a+c}+1=\frac{2c}{a+b}+1\)

=> \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}\)

^_^ 

Bài 1: Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\)

\(\Rightarrow\hept{\begin{cases}a=2016k\\b=2017k\\c=2018k\end{cases}}\).Thay vào M,ta có:

 \(M=4\left(2016k-2017k\right)\left(2017k-2018k\right)-\left(2018k-2016k\right)^2\)

\(=4.\left(-1k\right)\left(-1k\right)-\left(2k\right)^2\)

\(=4k^2-4k^2=0\)

Câu 2:   A =    \(^{1+2+2^2+2^{ }^3+...+2^{2017}}\)

          2A = \(2+2^2+2^3+...+2^{2018}\)

Suy ra 2A - A =\(2^{2018}-1\) Do đó A < B

1. Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=t\Rightarrow a=2016t,b=2017t,c=2018t\)

\(\left(a-c\right)^3=\left(2016t-2018t\right)^3=\left(-2t\right)^3=-8t^3\)

\(8\left(a-b\right)^2\left(b-c\right)=8\left(2016t-2017t\right)^2\left(2017t-2018t\right)=8.\left(-t\right)^2.\left(-t\right)=-8t^3\)

Vậy \(\left(a-c\right)^3=8\left(a-b\right)^2\left(b-c\right)\)

1) ADTCDTSBN

có: \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-7}=\frac{x-y-z}{3-5+7}=\frac{20}{5}=4.\)

=> ...

C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!