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Đặt dãy tỷ số bằng nhau là (1)
\(\Rightarrow\left(1\right)=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
\(\Rightarrow\left(1\right)=\frac{2\left(a+b\right)+3\left(c+d\right)}{c+d}=\frac{2\left(a+b\right)}{c+d}+3=5\Rightarrow\frac{\left(a+b\right)}{c+d}=1\)
Chứng minh tương tự ta tính và suy ra
\(\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
Từ giả thiết suy ra:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
* Nếu a + b + c + d = 0 thì a + b = - ( c + d ); b + c = - ( d + a ); c + d = - ( a + b ); d + a = - ( b + c )
Khi đó M = ( - 1 ) + ( - 1 ) + ( - 1 ) + ( - 1 ) = - 4
* Nếu a + b + c + d \(\ne0\) thì \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\)nên a = b = c = d
Khi đó M = 1 + 1 + 1 + 1 = 4
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
TH1: Nếu a+b+c+d\(\ne\)0 thì theo tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}\)\(=\frac{5a+5b+5c+5d}{a+b+c+d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
<=> \(2a+b+c+d=5a;a+2b+c+d=5b;a+b+2c+d=5c;a+b+c+2d=5d\)
<=>\(b+c+d=3a;a+c+d=3b;a+b+d=3c;a+b+c=3d\)
=>\(b+c+d+a+c+d=3a+3b\Leftrightarrow a+b+2c+2d=3a+3b\)
<=>\(2c+2d=2a+2b\Leftrightarrow2\left(c+d\right)=2\left(a+b\right)\Leftrightarrow c+d=a+b\)
Chứng minh tương tự ta được b+c=d+a ; c+d=a+b ; d+a=b+c
=>\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
TH2: a+b+c+d=0
\(\Leftrightarrow a+b=-\left(c+d\right);b+c=-\left(a+b\right);c+d=-\left(a+b\right);d+a=-\left(b+c\right)\)
\(\Rightarrow M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Vậy ........................
Trừ 1 ở mỗi phân số ta đuợc :
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(=\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu : a+b+c+d\(\ne\)0
=> a=b=c=d
=> \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
Nếu a+b+c+d=0
=> +) a+b=-(c+a)
+) b+c=-(d+a)
+) c+d=-(a+b)
+) d+a=-(b+c)
=> M=(-1)+(-1)+(-1)+(-1)=-4
Xét a+b+c+d = 0 ta có :
\(a+b=-c-d;b+c=-d-a;c+d=-a-b;d+a=-b-c\)
\(\Rightarrow M=\frac{-c-d}{c+d}+\frac{-d-a}{d+a}+\frac{-a-b}{a+b}+\frac{-b-c}{b+c}=-4\)
Xét a+b+c+d \(\ne0\) ta có :
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Leftrightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Leftrightarrow\frac{a+b+c+d}{a}=\frac{a+b+d+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)Thay vào M ta được : \(M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=4\)
OK:
Trừ 1 ở mỗi tỉ số,ta có:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1\)\(=\frac{a+b+c+2d}{d}-1\)
=>\(\frac{2a+b+c+d-a}{a}=\frac{a+2b+c+d-b}{b}\)\(=\frac{a+b+2c+d-c}{c}=\frac{a+b+c+2d-d}{d}\)
=>\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Do đó a=b=c=d
=>\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\)\(\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)
Vậy M=4
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 \(n\in\left\{-8;-2;0;6\right\}\)
còn ai nữa à =='
đk a,b,c,d khác 0
áp dugnj tc dãy tỉ số = nhau \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}\)
+> nếu a+b+c+d =0\(\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\end{cases}\hept{\begin{cases}d+a=-\left(b+c\right)\\\end{cases}}}\)\(\Rightarrow M=-4\)
+> a+b+c+d khác 0 \(\Rightarrow\frac{2a+b+c+d}{a}=5\Rightarrow b+c+d=3a\)
Tương tự ta có \(\hept{\begin{cases}a+b+c=3d\\a+c+d=3b\\a+b+d=3c\end{cases}}\)\(\Rightarrow a=b=c=d\)
Khi đó M=4
Vậy M=4 hoặc M=-4
cố lên 2 bác nha!!!