1) Ta có : \(\frac{2016a+b+c+d}{a}=\frac{a+2016b+c+d}{b}=\frac{a+b+2016c+d}{c}=\frac{a+b+c+2016d}{d}\)

Trừ 4 vế với 2015 ta được : \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

Nếu a + b + c + d = 0

=> a + b = -(c + d)

=> b + c = (-a + d) 

=> c + d = -(a + b)

=> d + a = (-b + c)

Khi đó M = (-1) + (-1) + (-1) + (-1) = - 4

Nếu a + b + c + d\(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)

Khi đó M = 1 + 1 + 1 + 1 = 4

2) a) Ta có : \(\hept{\begin{cases}\left|x+2013\right|\ge0\forall x\\\left(3x-7\right)^{2004}\ge0\forall y\end{cases}\Rightarrow\left|x+2013\right|+\left(3x-7\right)^{2014}\ge0}\)

Dấu "=" xảy ra \(\hept{\begin{cases}x+2013=0\\3y-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2013\\y=\frac{7}{3}\end{cases}}}\)

b) 7^2x + 7^{2x + 3} = 344

=> 7^2x + 7^2x.7³ = 344

=> 7^2x.(1 + 7³) = 344

=> 7^2x  = 1

=> 7^2x = 7⁰

=> 2x = 0 => x = 0

c) Ta có :

 \(\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{5}{x+4}\Leftrightarrow\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{10}{2x+8}=\frac{7-10}{2x+2-2x-8}=\frac{1}{2}\)(dãy tỉ số bằng nhau)

=>  2x + 2 = 14 => x = 6 ; 

2y - 4 = 6 => y = 5 ; 

6 + 5 + z = 17 => z = 6 

Vậy x = 6 ; y = 5 ; z = 6

3) a) Ta có : \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)(dãy ti số bằng nhau) 

=> a + b + c = a + b - c => a + b + c - a - b + c = 0 => 2c = 0 => c = 0;  

Lại có : \(\frac{a+b+c}{a+b-c}-1=\frac{a-b+c}{a-b-c}-1\Leftrightarrow\frac{2c}{a+b-c}=\frac{2c}{a-b-c}\Rightarrow a+b-c=a-b-c\) => b = 0 

Vậy c = 0 hoặc b = 0

c) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b+b+c+a+c}{c+a+b}=2\)(dãy tỉ số bằng nhau) 

=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)

Khi đó P = \(\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{b}{a}\right)=\frac{b+c}{b}.\frac{c+a}{c}=\frac{a+b}{a}=\frac{2a.2b.2c}{abc}=8\)

Vậy P = 8

2. b) \(7^{2x}+7^{2x+3}=344\)

        \(7^{2x}\cdot\left(1+7^3\right)=344\)

        \(7^{2x}\cdot\left(1+343\right)=344\)

        \(7^{2x}\cdot344=344\)

               \(7^{2x}=1\)  

               \(7^{2x}=7^0\)

              \(2x=0\)

               \(x=0\)

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Đặt \(\frac{a}{2}=\frac{c}{4}=\frac{e}{5}\) = k => a = 2k; c = 4k ; e = 5k

\(\frac{b}{3}=\frac{d}{5}=\frac{g}{6}\)= h => b = 3h; d = 5h; g = 6h

Khi đó: \(\frac{a}{b}+\frac{c}{d}+\frac{e}{g}=\frac{2k}{3h}+\frac{4k}{5h}+\frac{5k}{6h}=\left(\frac{2}{3}+\frac{4}{5}+\frac{5}{6}\right).\frac{k}{h}=2\frac{3}{10}\)

=> \(\frac{23}{10}.\frac{k}{h}=\frac{23}{10}\)=> \(\frac{k}{h}=1\)=> k = h

Vậy \(\frac{a}{b}=\frac{2k}{3h}=\frac{2}{3};\frac{c}{d}=\frac{4}{5};\frac{e}{g}=\frac{5}{6}\)