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24 tháng 6 2017
Áp dụng dãy tỉ số bằng nhau :
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}=\dfrac{a+b+c+d}{b+c+d+a+c+d+a+b+d+b+c+a}=\dfrac{1}{3}\) \(\Rightarrow3a=b+c+d\left(1\right)\)
\(\Rightarrow3b=c+d+a\left(2\right)\)
\(\Rightarrow3c=a+b+d\left(3\right)\)
\(\Rightarrow3d=b+c+a\left(4\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow3a+3b=b+c+d+c+d+a\)
\(\Rightarrow2a+2b=2c+2d\)
\(\Rightarrow a+b=c+d\)
Từ \(\left(2\right)+\left(3\right)\Rightarrow3b+3c=a+c+d+a+b+c\)
\(\Rightarrow2b+2c=2d+2a\)
\(\Rightarrow b+c=d+a\)
Từ \(\left(1\right)+\left(3\right)\Rightarrow2a+2c=2b+2d\)
\(\Rightarrow a+c=b+d\)
Ta có :
\(b+c=a+d;a+c=b+d\)
\(\Rightarrow b+c+a+c=d+a+b+a\)
\(\Rightarrow a+b+2c=2a+a+d\)
\(\Rightarrow c=d\)
Lại có :
\(b+c=d+a;a+c=b+d\)
\(\Rightarrow b+c+b+d=d+a+a+c\)
\(\Rightarrow2b+c+d=2a+d+c\)
\(\Rightarrow a=b\)
Từ những điều trên ta thấy được :
\(\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}=1+1+1+1=4\)
NT
24 tháng 6 2017
Nguyễn Thanh Hằng Xét thiếu TH rồi bạn !!!
Ta có :
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\\ \Rightarrow\dfrac{a}{b+c+d}+1=\dfrac{b}{a+c+d}+1=\dfrac{c}{a+b+d}+1=\dfrac{d}{a+b+c}+1\\ \Rightarrow\dfrac{a+b+c+d}{b+c+d}=\dfrac{a+b+c+d}{a+c+d}=\dfrac{a+b+c+d}{a+b+d}=\dfrac{a+b+c+d}{a+b+c}\)
TH1: Nếu a+b+c+d#0
thì Đỗ Thu Trà giải giống bạn Nguyễn Thanh Hằng
Nếu a+b+c+d=0 =>a+b=-(c+d); b+c=-(a+d);c+d=-(a+b); a+d=-(b+c)
Thế những cái này vao biểu thức M thì M=-4
26 tháng 5 2022
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
nên \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\)
c: \(\dfrac{a}{a+c}=\dfrac{bk}{bk+dk}=\dfrac{b}{b+d}\)
5 tháng 4 2017
Giải:
Ta có: \(\dfrac{2012a+b+c+d}{a}=\dfrac{a+2012b+c+d}{b}=\dfrac{a+b+2012c+d}{c}\)
\(=\dfrac{a+b+c+2012d}{d}\)
\(\Rightarrow\dfrac{2012a+b+c+d}{a}-2011=\dfrac{a+2012b+c+d}{b}-2011\)
\(=\dfrac{a+b+2012c+d}{c}-2011=\dfrac{a+b+c+2012d}{d}-2011\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
+) Xét \(a+b+c+d=0\) ta có:
\(\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(M=\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}\)
\(\Rightarrow M=\dfrac{-\left(c+d\right)}{c+d}=\dfrac{-\left(a+d\right)}{a+d}=\dfrac{-\left(a+b\right)}{a+b}=\dfrac{-\left(b+c\right)}{b+c}=-1\)
+) Xét \(a+b+c+d\ne0\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}\)
\(\Rightarrow M=\dfrac{2a}{2a}=\dfrac{2a}{2a}=\dfrac{2a}{2a}=\dfrac{2a}{2a}=1\)
Vậy nếu \(a+b+c+d=0\) thì M = -1
nếu \(a+b+c+d\ne0\) thì M = 1
31 tháng 5 2017
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Lần lượt thay a và c vào các ý cần chứng minh, áp dụng theo tính chất phân phối giữa phép nhân đối với phép cộng (hay phép trừ) để tính ở mỗi vế.
Mẫu: a) Ta có : \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Vậy \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
LL
13 tháng 8 2017
a)\(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Gọi\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(a=b.k\)
\(c=d.k\)
\(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b.\left(k+1\right)}{b}=k+1\) (1)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d.\left(k+1\right)}{d}=k+1\)(2)
Từ (1) và (2) \(\Rightarrow\)\(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b)\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
Gọi\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(a=b.k\)
\(c=d.k\)\(\dfrac{a-b}{a}=1-\dfrac{b}{a}=1-\dfrac{b}{bk}=1-\dfrac{1}{k}\left(1\right)\)
\(\dfrac{c-d}{c}=1-\dfrac{d}{c}=1-\dfrac{d}{dk}=1-\dfrac{1}{k}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\)\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
5 tháng 12 2017
ta có;\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)\(=>\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)\(=>\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}=\dfrac{a+b+c+d+a+b+c+d+a+b+c+d+a+b+c+d}{a+b+c+d}=\dfrac{4.\left(a+b+c+d\right)}{a+b+c+d}=4\)=>\(\left\{{}\begin{matrix}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\\a+b+c+d=4d\end{matrix}\right.\)
Nếu a=b=c=d=0=>\(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+a}{a+b}+\dfrac{a+a}{b+c}=0\)
Nếu a,b,c,d≠0=>4a=4b=4c=4d
=>a=b=c=d
do đó;\(\dfrac{a+a}{a+a}+\dfrac{b+b}{b+b}+\dfrac{c+c}{c+c}+\dfrac{d+d}{d+d}=1+1+1+1=4\)
5 tháng 12 2017
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
Ta sẽ chứng minh phương trình sau chỉ đúng khi: \(\left[{}\begin{matrix}a+b+c+d=0\\a=b=c=d\end{matrix}\right.\)
Thật vậy:
Từ: \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}\Leftrightarrow a\left(a+b+c+d\right)=b\left(a+b+c+d\right)\)\(\Leftrightarrow\left(a-b\right)\left(a+b+c+d\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b+c+d=0\end{matrix}\right.\)(1)
Từ: \(\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}\Leftrightarrow b\left(a+b+c+d\right)=c\left(a+b+c+d\right)\Leftrightarrow\left(b-c\right)\left(a+b+c+d\right)\Leftrightarrow\left[{}\begin{matrix}b=c\\a+b+c+d=0\end{matrix}\right.\)(2)
Từ: \(\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\Leftrightarrow c\left(a+b+c+d\right)=d\left(a+b+c+d\right)\Leftrightarrow\left(c-d\right)\left(a+b+c+d\right)=0\Leftrightarrow\left[{}\begin{matrix}c=d\\a+b+c+d=0\end{matrix}\right.\)(3)
Phương trình cần chứng minh cần thỏa mãn cả 3 phương trình (1);(2);(3),hay \(\left[{}\begin{matrix}a+b+c+d=0\\a=b=c=d\end{matrix}\right.\)
\(\circledast\) Với \(a+b+c+d=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\Leftrightarrow A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
\(\circledast\) Với \(a=b=c=d\Leftrightarrow P=1+1+1+1=4\)
Vậy \(A=\left[{}\begin{matrix}4\\-4\end{matrix}\right.\)
6 tháng 11 2017
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}\dfrac{1}{3}\)(vìa+b+c+d\(\ne\)0)
=>3a=b+c+d: 3b=a+c+d=>3a-3b=b-a
=>3(a-b)=-(a-b)=>4(a-b)=0=>a=b
Tương tự => a=b=c=d=> A=4
3 tháng 2 2018
Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{a+b}{a+b+2\left(c+d\right)}=\dfrac{1}{3}\)
\(\Rightarrow3\left(a+b\right)=\left(a+b\right)+2\left(c+d\right)\)
\(\Rightarrow2\left(a+b\right)=2\left(c+d\right)\)
\(\Rightarrow a+b=c+d\)
\(\Rightarrow\dfrac{a+b}{c+d}=1\)
Tương tự:\(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
Vậy A=4.
TM
10 tháng 1 2018
Ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
+) Nếu \(a+b+c+d\ne0\) thì từ trên suy ra:\(a=b=c=d\)
\(\Rightarrow M=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
+) Nếu \(a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow M=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=\left(-4\right)\)
Vậy M = 4 hoặc M = -4