cho: 2a+b+c+d/a = a+2b+c+d/b = a+b+2c+d/c = a+b+c+2d/d
tính: (a+b/c+d )+ (b+c/d+a) +( c+d/a+b) + (d+a/b+c)
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XO
26 tháng 10 2020
Ta có\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
=> \(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
=> \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Khi a + b + c + d = 0
=> a + b = -(c + d)
b + c = -(a + d)
Khi đó \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{a+d}{b+c}\)
\(=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{c+d}{-\left(c+d\right)}+\frac{a+d}{-\left(a+d\right)}=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)= -4
Nếu a + b + d + d \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)
Khi đó M = \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{2a}{2a}+\frac{2b}{2b}+\frac{2c}{2c}+\frac{2d}{2d}=1+1+1+1=4\)
Vậy khi a + b + c + d = 0 => M = -4
khi a + b + c + d \(\ne\)0 => M = 4
23 tháng 10 2015
Cho biểu thức sau:$\frac{2a+b+c+d}{a}$2 a + b + c + d a bam vao do nho bam lik e :\
XO
20 tháng 10 2019
Theo bài ra ta có :
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a + b + c + d = 0
\(\Rightarrow\frac{0}{a}=\frac{0}{b}=\frac{0}{c}=\frac{0}{d}\)
\(\Rightarrow\orbr{\begin{cases}a=b=c=d\\a\ne b\ne c\ne d\end{cases}}\)(loại)
Nếu a + b + c + d \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\)
=> a = b = c = d (đpcm)
NH
1
28 tháng 3 2019
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\left(\text{ vì a+b+c+d khác 0}\right)\)
\(\Rightarrow a=b=c=d\)
\(M=\frac{2a-b}{c+b}+\frac{2b-c}{a+d}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2b-b}{b+b}+\frac{2c-c}{c+c}+\frac{2d-d}{d+d}=\frac{1}{2}.4=2\)
LL
1
* TH1: a + b + c + d \(\ne\)0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}\)
\(=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
\(\Rightarrow2a+b+c+d=5a;a+2b+c+d=5b\)
\(\Rightarrow b+c+d=3a;a+c+d=3b\)
\(\Rightarrow b+c+d+a+c+d=3a+3b\)
\(\Rightarrow\left(a+b\right)+2\left(c+d\right)=3\left(a+b\right)\)
\(\Rightarrow2\left(c+d\right)=2\left(a+b\right)\)
\(\Rightarrow c+d=a+b\)
CMTT ta được: \(b+c=a+d\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
* TH2: \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right);b+c=-\left(d+a\right)\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)\(=-4\)
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