Ta có: \(\hept{\begin{cases}VT=\left(a+2c\right)\left(b+d\right)=ab+ad+2bc+2cd\\VP=\left(a+c\right)\left(b+2d\right)=ab+2ad+bc+2cd\end{cases}}\)

Từ \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow VT=VP\Leftrightarrowđpcm\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\\ =>\orbr{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(Taco:\left(a+2c\right).\left(b+d\right)=\left(a+c\right).\left(b+2d\right)\)

\(=>\left(bk+2dk\right).\left(b+d\right)=\left(bk+dk\right).\left(b+2d\right)\)

\(=>\frac{bk+2dk}{bk+dk}=\frac{b+2d}{b+d}\)

\(=>\frac{k.\left(b+2d\right)}{k.\left(b+d\right)}=\frac{b+2d}{b+d}\)

\(=>\frac{b+2d}{b+d}=\frac{b+2d}{b+d}\)(ĐPCM)

, Chờ tí mk làm câu b

        Ta có :\(\frac{a}{b}=\frac{c}{d}\)

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

Từ (1);(2)\(​​​\implies\) \(\frac{a+2c}{b+2d}=\frac{a+c}{b+d}\)  

                 \(\implies\) \(\left(a+2c\right).\left(b+d\right)=\left(b+2d\right).\left(a+c\right)\)

P/S : ko chắc 

Áp dụng tc của dãy tỉ số bằng nhau có :

\(\frac{a}{b}=\frac{c}{d}=\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)(ĐPCM)

Đánh máy ẩu v :D

thôi ko cần nx đâu,mình làm được rồi,cảm ơn các bạn nha!!!

khó quá ak

ừ, bạn bik làm thì giúp mình nha ^^

I DON NO

a/b = c/d

=> a = bk và c = dk

thay vào ta có : 

(bk + 2dk)(b+d) = (bk+dk)(b+2d)

=> k(b+2d)(b+d) = k(b+d)(b+2d)

xong rồi nha

K bt thì k cần phải nói 

a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)

Thay:

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

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

=> đpcm

a, \(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\Rightarrow\frac{a^4}{b^4}=\frac{c^4}{d^4}=\frac{\left(a-c\right)^4}{\left(b-d\right)^4}\) (1)

\(\frac{a^4}{b^4}=\frac{c^4}{d^4}=\frac{5a^4}{5b^4}=\frac{7c^4}{7d^4}=\frac{5a^4+7c^4}{5b^4+7d^4}\)(2)

Từ (1) và (2) => đpcm

b, \(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+2c}{b+2d}\) (3)

\(\frac{a}{b}=\frac{c}{d}=\frac{3c}{3d}=\frac{a-3c}{b-3d}\) (4)

Từ (3) và (4) => đpcm

c, làm giống câu a

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

            \(\frac{a}{b}=\frac{c}{d}=\frac{a-3c}{b-3d}\left(2\right)\)

(1) và (2) => \(\frac{a+2c}{b+2d}=\frac{a-3c}{b-3d}\)

Tương tự \(\left(\frac{a}{b}\right)^4=\left(\frac{c}{d}\right)^4=\left(\frac{a-c}{b-d}\right)^4\left(1\right)\)

               \(\left(\frac{a}{b}\right)^4=\left(\frac{c}{d}\right)^4=\frac{5a^4+7c^4}{5b^4+7d^4}\left(2\right)\)

=> \(\left(\frac{a-c}{b-d}\right)^4=\frac{5a^4+7c^4}{5b^4+7d^4}\)

Đặt

\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(bk+2dk\right)\left(b+d\right)\)

\(=bk\left(b+d\right)+2dk\left(b+d\right)\)

\(=b^2k+bdk+2bdk+2d^2k\)

\(=b^2k+3bdk+2d^2k\)

\(\Rightarrow\left(a+c\right)\left(b+2d\right)=\left(bk+dk\right)\left(b+2d\right)\)

\(=bk\left(b+2d\right)+dk\left(b+2d\right)\)

\(=b^2k+2bdk+bdk+2d^2k\)

\(=b^2k+3bdk+2d^2k\)

\(VT=VP\)\(\Rightarrowđpcm\)