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\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}=\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Rightarrow a=b=c\Rightarrow M=1\)
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) 72x + 72x + 3 = 344
=> 72x + 72x.73 = 344
=> 72x.(1 + 73) = 344
=> 72x = 1
=> 72x = 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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
\(\frac{a}{b}=\frac{14}{22}=\frac{7}{11}\Rightarrow\frac{a}{7}=\frac{b}{11}=\frac{a+b}{7+11}=\frac{M}{18}\)
\(\frac{c}{d}=\frac{11}{13}\Rightarrow\frac{c}{11}=\frac{d}{13}=\frac{c+d}{11+13}=\frac{M}{24}\)
\(\frac{e}{f}=\frac{13}{17}\Rightarrow\frac{e}{13}=\frac{f}{17}=\frac{e+f}{13+17}=\frac{M}{30}\)
Mà M là số tự nhiên => M là bội chung nhỏ nhất của 18; 24; 30
18 = 32.2 ; 24 = 3.23 ; 30 = 2.3.5
=> BCNN (18;24;30 ) = 32.23.5 = 360
Hay M = 360
a) \(\frac{-3}{5}.y=\frac{21}{10}\)
\(y=\frac{21}{10}:\frac{-3}{5}\)
\(y=\frac{-7}{2}\)
vậy \(y=\frac{-7}{2}\)
b) \(y:\frac{3}{8}=-1\frac{31}{33}\)
\(y:\frac{3}{8}=\frac{-64}{33}\)
\(y=\frac{-64}{33}.\frac{3}{8}\)
\(y=\frac{-8}{11}\)
vậy \(y=\frac{-8}{11}\)
c) \(1\frac{2}{5}.y+\frac{3}{7}=\frac{-4}{5}\)
\(\frac{7}{5}.y+\frac{3}{7}=\frac{-4}{5}\)
\(\frac{7}{5}.y=\frac{-4}{5}-\frac{3}{7}\)
\(\frac{7}{5}.y=\frac{-43}{35}\)
\(y=\frac{-43}{35}:\frac{7}{5}\)
\(y=\frac{-43}{49}\)
vậy \(y=\frac{-43}{49}\)
d) \(\frac{-11}{12}.y+0,25=\frac{5}{6}\)
\(\frac{-11}{12}.y=\frac{5}{6}-0,25\)
\(\frac{-11}{12}.y=\frac{7}{12}\)
\(y=\frac{7}{12}:\frac{-11}{12}\)
\(y=\frac{-7}{11}\)
vậy \(y=\frac{-7}{11}\)
72;120;210;385