Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)

\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)

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

b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)

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

Đặt\(a+c=2b\left(1\right);2bd=c\left(b+d\right)\left(2\right)\\ \)

Thay (1) vào (2):\(\left(a+c\right)d=c\left(b+d\right)\)

Khai triển hết ra r rút gọn là ok.

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

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 \(k=-1\) :

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

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

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

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

\(\Rightarrow P=4\)

Bài 1 :

Ta có :

\(a+c=2b\left(1\right)\)

\(2bd=c\left(b+d\right)\left(2\right)\)

Thay \(\left(1\right)\) vào \(\left(2\right)\) ta được :

\(\left(a+c\right)d=c\left(b+d\right)\)

\(\Leftrightarrow ad+cd=cb+cd\)

\(\Leftrightarrow ad=cb\)

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

Bài 2 :

\(a,\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)

\(=\dfrac{5^{21}\left(2.5-9\right)}{5^{20}}\)

\(=5\left(10-9\right)\)

\(=5\)

b, \(\dfrac{5\left(3.7^{15}-19.17^{14}\right)}{7^{14}+3.7^{15}}\)

\(=\dfrac{5.2.7^{14}}{10.7^{15}}\)

\(=\dfrac{1}{7}\)

câu hỏi tương tự nha bạn.

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

Ta có VT:

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)

\(=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\) (1)

VT: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\) (2)

Từ (1) và (2) 

\(\Rightarrow\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\left(đpcm\right)\)

Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ab=cd\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

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

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

Vậy...

\(a+c=2b\\ \Leftrightarrow d\left(a+c\right)=2bd\\\Leftrightarrow d\left(a+c\right)=c\left(b+d\right) \\ \Leftrightarrow ad+cd=cb+cd\\ \Leftrightarrow ad=cb\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

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