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) \(7^{2x}+7^{2x+3}=344\)

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

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

        \(7^{2x}\cdot344=344\)

               \(7^{2x}=1\)  

               \(7^{2x}=7^0\)

              \(2x=0\)

               \(x=0\)

\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{100}-1\right)\)

\(-A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{100}\right)\)

\(-A=\frac{9}{10}\cdot\frac{10}{11}\cdot\frac{11}{12}\cdot...\cdot\frac{99}{100}\)

\(-a=\frac{9}{100}\)

\(A=-\frac{9}{100}\)

Bài 1.

Ta có: \(\frac{a}{b}+\frac{-a}{b+1}=\frac{a}{b}-\frac{a}{b+1}=a\left(\frac{1}{b}-\frac{1}{b+1}\right)=a\left(\frac{b+1-b}{b\left(b+1\right)}\right)=a\left(\frac{1}{b\left(b+1\right)}\right)=\frac{a}{b\left(b+1\right)}\)

=> A là đáp án đúng

Bài 2. Ta có:

B = 4x - 4y + 5xy

B= 4x - 4y + 4xy + xy

B = 4(x - y + xy) + xy

B = 4.(5/12 - 1/3) - 1/3

B = 4.1/12 - 1/3
B = 1/3 - 1/3 = 0

a/ \(\left(3x-1\right)^6=\left(3x-1\right)^4\Rightarrow\left(3x-1\right)=\left\{-1;0;1\right\}\)

\(\Rightarrow x=\left\{0;\frac{1}{3};\frac{2}{3}\right\}\)

b/

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

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

Tương tự

\(b+c=2a;a+c=2b\)

\(\Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2c.2a.2b}{abc}=8\)

Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1

        c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1

=> A = 1+bc+b/bc+b+1 = 1

Tk mk nha

BÀI 1:

\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)        

\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\)       (thay   abc = 1)

\(=\frac{a+ab+1}{a+ab+1}=1\)

Bài làm:

Áp dụng t/c dãy tỉ số bằng nhau:

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{c+a+b}\)

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

\(\Rightarrow\hept{\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b+c=3c\\a+b+c=3a\\a+b+c=3b\end{cases}}\Rightarrow a=b=c\)

Thay vào ta tính được:

\(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)

\(B=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2^3=8\)

Vậy B = 8

Ta có : \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

\(\Rightarrow\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

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

Nếu a + b + c = 0

=> a + b = -c

=> a + c = -b

=> b + c = -a

Khi đó B = \(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=-\frac{abc}{abc}=-1\)

Nếu a + b + c \(\ne\)0

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

Khi đó B = \(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)

Vậy khi a + b + c = 0 => B = -1

khi a + b + c \(\ne\)0 => B = 8

e) kq=-5