Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?

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

vì -1 hơn 1 hai số cho nên;

a) a/b và c/d ^2 =ab/cd hơn kém nhau 2

b) dựa theo tính chất kết hợp (a+b/c+d ) ^3 = a ^3 ...

quá đơn giản 

cho 5 k giải cho

(mình trong đội tuyển toán đó nhe nên làm theo đi)

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

=> cd(a² + b²) = ab(c² + d²) 

=> a²cd + b²cd = abc² + abd²

=>  a²cd + b²cd - abc² - abd² = 0

=>  (a²cd - abc²) + (b²cd - abd²) = 0

=> ac(ad - bc) + bd(bc - ad) = 0

=> ac(ad - bc) - bd(ad - bc) = 0

=> (ac - bd)(ad - bc) = 0

=> \(\orbr{\begin{cases}ac-bd=0\\ad-bc=0\end{cases}}\Rightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{d}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\Rightarrow\text{đpcm}\)

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

\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\\\dfrac{a}{c}=\dfrac{b}{d}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(\dfrac{a}{c}\right)^2=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\\\left(\dfrac{a}{c}\right)^2=\dfrac{ab}{cd}\end{matrix}\right.\)

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

Ta có : a^2+b^2/c^2+d^2 = ab/cd 
=> (a^2+b^2) . cd = (c^2+d^2). ab 
=> a.a.c.d+b.b.c.d = c.c.a.b + d.d.a.b
=> a.a.c.d-c.c.a.b - d.d.a.b + b.b.c.d= 0
=> ac(ad - bc) - bd(ad - bc) = 0 
=> (ac - bd)(ad - bc) = 0 
=> ac - bd = 0 hoặc ad - bc = 0 
=> ac = bd 
=> a/b =c/d  (đpcm)
 

Ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)

\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)

\(\Leftrightarrow\left(a^2cd-abd^2\right)+\left(b^2cd-abc^2\right)=0\)

\(\Leftrightarrow ad\left(ac-bd\right)-bc\left(ac-bd\right)=0\)

\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\) (đpcm)