Question

cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}\left(a+b+c+d\ne0\right)\). tính \(P=\dfrac{2a-b}{c+d}+\dfrac{2b-c}{a+đ}+\dfrac{2c-d}{a+b}+\dfrac{2c-a}{b+c}\)

Answer 1

Vì \(a+b+c+d\ne0\) nên áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

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

\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\) \(\Rightarrow a=b=c=d\) (1)

Thay (1) vào P, ta có:

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

\(\Rightarrow P=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=2\)

Vậy P = 2

Answer 2

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=k\)

\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{a}=k^4\)

\(\Rightarrow k=\pm1\)

- Với \(k=1\) :

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

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

- Với \(k=-1\) :

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

\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-d\\d=-a\end{matrix}\right.\)

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

\(\Rightarrow P=\dfrac{2a+a}{2a+a}+\dfrac{-2a-a}{-2a-a}+\dfrac{2a+a}{2a+a}+\dfrac{-2a-a}{-2a-a}\)

\(\Rightarrow P=4\)