Theo đề bài ta có :

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

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

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

Theo tính chất dãy tỉ số bằng nhau ta có :

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

\(k^2=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)  ( 2 )

Mà từ ( 1 ) = > \(k^2=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\) ( 3 )

Từ ( 2 ) , ( 3 ) 

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)

Ta có: \(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\left(1\right)\)

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

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\Rightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\\ \dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Cách giải:

1+1=3

6-6=0

9-9=0

Vậy => 6-6=9-9

(3-3)+(3-3) = 3x3 - 3x3

(1+1)=3

1+1=3

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 \(\dfrac{a}{b}=\dfrac{d}{c}\) ?

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

=> \(\frac{a}{c}=\frac{b}{d}\)

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)(Tính chất dãy tỉ số bằng nhau)

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

 a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)

hey you, còn câu b,c?

ở đây có ai thích sơn tùng không ?

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

không cs số 0 đâu 

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Sửa: \(\dfrac{3a^2+10b^2-ab}{7a^2+b^2+5ab}=\dfrac{3b^2k^2+10b^2-b^2k}{7b^2k^2+b^2+5b^2k}=\dfrac{b^2\left(3k^2+10-k\right)}{b^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(1\right)\)

\(\dfrac{3c^2+10d^2-cd}{7c^2+d^2+5cd}=\dfrac{3d^2k^2+10d^2-d^2k}{7d^2k^2+d^2+5d^2k}=\dfrac{d^2\left(3k^2+10-k\right)}{d^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(2\right)\)

\(\left(1\right)\left(2\right)\RightarrowĐpcm\)