Ta có :

\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)

\(\frac{a^2+b^2}{c^2+d^2}-\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)

Từ ( 1 ) và ( 2 ) suy ra : \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)

TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{b}\left(3\right)\)

        \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)

từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\text{ hay }\frac{a}{b}=\frac{c}{d}\)

TH2 : \(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)

     \(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)

Từ ( 5 ) và ( 6 ) suy ra : \(\frac{b}{c}=\frac{a}{d}\text{ hay }\frac{a}{b}=\frac{d}{c}\)

Vậy : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\text{ thì }\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)

kinh quá

Xin lỗi, mình bị nhầm, bài này của lớp 8.

Xin lỗi, mình bị nhầm, bài này của lớp 8

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\)\(\frac{d}{a+b+c}\)

\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)

\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)

Mà: \(a+b+c+d\ne0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)

\(\Rightarrow a=b=c=d\)

\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)

\(\Rightarrow A=1+1+1+1=4\)

số đo slaf

4 

nhe sbn

bài dài 

lắm mình

vhir tiện ghi

thế này thôi

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)

\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)

Mà :\(a+b+c+d=0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)

\(\Rightarrow a=b=c=d\)

\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)

\(\Rightarrow A=1+1+1+1=4\)

=2 tick mk nha

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)

\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}\)\(=\frac{a+b+c+d}{a+b+c}\)

Do a + b + c + d khác 0 nên: b+c+d = a+c+d = a+b+d = a+b+c  => a = b = c = d

\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)\(\left(a=b=c=d\right)\)

\(\Rightarrow A=1+1+1+1=4\)

Giải: Ta có : 

\(\frac{a+b+c-2011d}{d}=\frac{b+c+d-2011a}{a}=\frac{c+d+a-2011b}{b}=\frac{d+a+b-2011c}{c}\)

=> \(\frac{a+b+c}{d}-2011=\frac{b+c+d}{a}-2011=\frac{c+d+a}{b}-2011=\frac{d+a+b}{c}-2011\)

=> \(\frac{a+b+c}{d}=\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}\)

=> \(\frac{a+b+c}{d}+1=\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{d+a+b}{c}+1\)

=> \(\frac{a+b+c+d}{d}=\frac{b+c+d+a}{a}=\frac{c+d+a+b}{b}=\frac{d+a+b+c}{c}\)

TH1: a + b + c + d = 0

=> a + b = -(c + d)

    b + c = -(a + d)

 khi đó, ta có : S = \(\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(a+d\right)}\)

                          = \(-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

                          = -4

TH2 : a + b + c + d \(\ne\)0

=> a = b = c = d

khi đó, ta có : S = \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{b+a}+\frac{d+a}{b+c}\)

                          =   1 + 1 + 1 + 1 

                         = 4

Vì \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{a^{2009}}{b^{2009}}=\frac{c^{2009}}{d^{2009}}=\left(\frac{a}{b}\right)^{2009}=\frac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}\)( áp dụng tc của dãy tỉ số bằng nhau )

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