Ta có :

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

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

\(\Rightarrow\frac{2012a}{2012c}=\frac{2013b}{2013d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau :

\(\frac{2012a}{2012c}=\frac{2013b}{2013d}=\frac{2012a+2013b}{2012c+2013d}=\frac{2012a-2013b}{2012c-2013d}\)

\(\Rightarrow\frac{2012a+2013b}{2012a-2013b}=\frac{2012c+2013d}{2012c-2013d}\)

Vậy...

dễ vậy mà k nghỉ ra cam mơm

Giải:

Ta có : a/b = c/d => a/c = b/d

Đặt a/c = b/d = k => a = ck ; b = dk

Khi đó, ta có : \(\frac{2012.ck+2013.dk}{2012.ck-2013.dk}=\frac{\left(2012c+2013d\right).k}{\left(2012c-2013d\right).k}=\frac{2012c+2013d}{2012c-2013d}\)(đpcm)

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

\(\Rightarrow a=\frac{2b}{2}=b\)                       \(c=\frac{2d}{2}=d\)

\(b=\frac{2c}{2}=c\)                               \(d=\frac{2a}{2}=a\)

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

Ta có: \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)

\(=\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}\)

\(=\frac{4a}{2a}=2\)

A ₫ 2 day ban so yeoung cheing nhe. Cac ban kcho mik nha

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Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}.\dfrac{a}{b}=\dfrac{1}{2}.\dfrac{b}{c}=\dfrac{1}{2}.\dfrac{c}{d}=\dfrac{1}{2}.\dfrac{d}{a}\)

⇒  \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{b+c+d+a}=1\)

⇒   \(a=b=c=d\)

Thay b = a ; c = a ; d = a vào biểu thức A ta có:

\(A=\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}\)

\(A=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)

\(A=\dfrac{1}{2}.4=2\)

Vậy A = 2

\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2a+2b+2c+2d}=\dfrac{1}{2}\)

=>\(\dfrac{a}{2b}=\dfrac{1}{2}\)=>2a=2b =>a=b

\(\dfrac{b}{2c}=\dfrac{1}{2}\)=>2b=2c =>b=c

\(\dfrac{c}{2d}=\dfrac{1}{2}\)=>2c=2d =>c=d

\(\dfrac{d}{2a}=\dfrac{1}{2}\)=>2d=2a =>d=a

=>a=b=c=d.

*\(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)

=\(\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010d}{a+a}+\dfrac{2011a-2010a}{a+a}\)

=\(\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)=2

Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}\cdot\dfrac{a}{b}=\dfrac{1}{2}\cdot\dfrac{b}{c}=\dfrac{1}{2}\cdot\dfrac{c}{d}=\dfrac{1}{2}\cdot\dfrac{d}{a}\)

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

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

Thay \(b=a;c=a;d=a\) vào biểu thức A ta có;

\(A=\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}\)

\(A=\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}\)

\(A=\dfrac{1}{2}\cdot4=2\)

Vậy \(A=2\)