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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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
\(\frac{b+c+d}{a}\)= \(\frac{c+d+a}{b}\)= \(\frac{d+a+b}{c}\)= \(\frac{a+b+c}{d}\)
= \(\frac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)
= \(\frac{3a+3b+3c+3d}{a+b+c+d}\)
= \(\frac{3\left(a+b+c+d\right)}{a+b+c+d}\)= 3
vậy k = 3
b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=k
áp dụng tc dãy tỉ số bằng nhau ta được:
b+c+d+c+d+a+d+a+b+a+b+c/a+b+c+d=k
=>3a+3b+3c+3d/a+b+c+d=k
=>3+k
=>k=3
Vậy k=3
\(\frac{a}{b}+\frac{c}{d}=\frac{a}{b}\cdot\frac{c}{d}\)
đề như vậy á?
Cộng thêm 1 vào mỗi tỉ số đã cho ta được:
\(\frac{b+c+d}{a}\) +1 = \(\frac{c+d+a}{b}\) +1 = \(\frac{d+a+b}{c}\) +1= \(\frac{a+b+c}{d}\) +1
\(\frac{a+b+c+d}{a}\) = \(\frac{a+b+c+d}{b}\) = \(\frac{a+b+c+d}{c}\) = \(\frac{a+b+c+d}{d}\)
Vì a+b+c+d khác 0 nên a=b=c=d
Suy ra k= \(\frac{3a}{a}\) = 3
1. \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1\)\(=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(=\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)(1)
TH1: \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right)\); \(b+c=-\left(d+a\right)\); \(c+d=-\left(a+b\right)\); \(d+a=-\left(b+c\right)\)
\(\Rightarrow M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+2017=-4+2017=2013\)
TH2: \(a+b+c+d\ne0\)
Từ (1) \(\Rightarrow a=b=c=d\)\(\Rightarrow M=1+1+1+1+2017=4+2017=2021\)
Vậy \(M=2013\)hoặc \(M=2021\)
2. \(2n-5=2n+2-7=2\left(n+1\right)-7\)
Vì \(2\left(n+1\right)⋮n+1\)\(\Rightarrow\)Để \(2n-5⋮n+1\)thì \(7⋮n+1\)
\(\Rightarrow n+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)\(\Rightarrow n\in\left\{-8;-2;0;6\right\}\)
Vậy \(n\in\left\{-8;-2;0;6\right\}\)
Bài 1:
Theo đề, ta có:
\(\dfrac{-13}{2}< \dfrac{11}{a}< \dfrac{-13}{3}\)
\(\Leftrightarrow\dfrac{-143}{26}< \dfrac{-143}{-13a}< \dfrac{-143}{33}\)
\(\Leftrightarrow\dfrac{143}{26}>\dfrac{143}{-13a}>\dfrac{143}{33}\)
hay \(a\in\varnothing\)
\(\frac{a}{b}+\frac{c}{d}=\frac{a.c}{b.d}\)
Vậy \(\frac{a.d+b.c}{bd}=\frac{ac}{bd}\)
\(\Leftrightarrow ad+bc=ac\)
\(\Leftrightarrow ad=ac-bc\)
\(\Leftrightarrow ad=c\left(a-b\right)\)
\(\Leftrightarrow\frac{a}{b}=\frac{a}{a-b}\)
\(\frac{a}{b}+\frac{c}{d}=\frac{a\cdot c}{b\cdot d}\)
Vậy \(\frac{a\cdot b+b\cdot c}{bd}=\frac{ac}{bd}\)
\(\Rightarrow ad+bc=ac\)
\(\Rightarrow ad=ac-bc\)
\(\Rightarrow ad=c\left(a-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a}{a-b}\)