Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}=\frac{1111c-99b}{9999c-11b}=\frac{1111a-99d}{9999a-11d}\)

\(\Rightarrow\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}\left(đpcm\right)\)

Vậy ....

a) \(\frac{4a-3b}{a}=\frac{4c-3d}{c}\Leftrightarrow4-\frac{3b}{a}=4-\frac{3d}{c}\)

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

b) \(\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}\Leftrightarrow\frac{9\left(9999-11d\right)-88880c}{9999c-11d}=\frac{9\left(9999a-11b\right)-88880a}{9999a-11b}\)

\(\Leftrightarrow9+\frac{-88880c}{9999c-11d}=9+\frac{-88880a}{9999a-11b}\)

\(\Leftrightarrow\frac{c}{9999c-11d}=\frac{a}{9999a-11b}\)

\(\Leftrightarrow\frac{9999c-11d}{c}=\frac{9999a-11b}{a}\)

\(\Leftrightarrow9999-\frac{11d}{c}=9999-\frac{11b}{a}\Leftrightarrow\frac{d}{c}=\frac{b}{a}\Leftrightarrow\frac{c}{d}=\frac{a}{b}\)

câu b hình như đề sai

ta có:

\(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)

\(\Rightarrow\frac{7a-11b}{7c-11d}=\frac{4a+5b}{4c+5d}\)

\(\Leftrightarrow\frac{7a}{7c}=\frac{11b}{11d}=\frac{4a}{4c}=\frac{5b}{5d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Mặt khác:

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

\(\Rightarrowđpcm\)

ta có:

Mặt khác:

khó quá ak

ừ, bạn bik làm thì giúp mình nha ^^

viết đề sai

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

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

\(\Rightarrow\frac{2a}{2c}=\frac{7b}{7d}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{2a}{2c}=\frac{7b}{7d}=\frac{2a+7b}{2c+7d}\) (1).

\(\frac{2a}{2c}=\frac{7b}{7d}=\frac{2a-7b}{2c-7d}\) (2).

Từ (1) và (2) \(\Rightarrow\frac{2a+7b}{2c+7d}=\frac{2a-7b}{2c-7d}.\)

\(\Rightarrow\frac{2a+7b}{2a-7b}=\frac{2c+7d}{2c-7d}\left(đpcm\right).\)

Chúc bạn học tốt!

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

Ta có: \(\frac{1111.c-99.d}{9999.c-11.d}=\frac{11.\left(101.c-9.d\right)}{11.\left(909.c-d\right)}=\frac{101.c-9.d}{909.c-d}=\frac{101.dk-9.d}{909.dk-d}=\frac{d.\left(101k-9\right)}{d.\left(909k-1\right)}=\frac{101k-9}{909k-1}\left(1\right)\)

\(\frac{1111.a-99.b}{9999.a-11.b}=\frac{11.\left(101a-9b\right)}{11.\left(909a-b\right)}=\frac{101a-9b}{909a-b}=\frac{101.bk-9b}{909.bk-b}=\frac{b.\left(101k-9\right)}{b.\left(909k-1\right)}=\frac{101k-9}{909k-1}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{1111.c-99.d}{9999.c-11.d}=\frac{1111.a-99.b}{9999.a-11.b}\left(đpcm\right)\)

Đặt \(k=\frac{a}{b}=\frac{c}{d}\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)

=> \(\frac{1111c-99d}{9999c-11d}=\frac{1111kd-99d}{9999kd-11d}=\frac{d\left(1111k-99\right)}{d\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\left(1\right)\)

\(\frac{1111a-99b}{9999a-11b}=\frac{1111kb-99b}{9999kb-11b}=\frac{b\left(1111k-99\right)}{b\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\left(2\right)\)

Từ (1) và (2) => \(\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}\)

a)Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)\(\Leftrightarrow\frac{bk-b}{b}=\frac{dk-d}{d}\)

Xét VT \(\frac{bk-b}{b}=\frac{b\left(k-1\right)}{b}=k-1\left(1\right)\)

Xét VP \(\frac{dk-d}{d}=\frac{d\left(k-1\right)}{d}=k-1\left(2\right)\)

Từ (1) và (2) =>Đpcm

b)Đặt tương tự ta xét VT:

\(\frac{11bk+3b}{11dk+3d}=\frac{b\left(11k+3\right)}{d\left(11k+3\right)}=\frac{b}{d}\left(1\right)\)

Xét VP \(\frac{3bk-11b}{3dk-11d}=\frac{b\left(3k-11\right)}{d\left(3k-11\right)}=\frac{b}{d}\left(2\right)\)

Từ (1) và (2) =>Đpcm

c)Cũng đặt tương tự

Xét VT \(\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)

Xét VP \(\frac{bk\cdot dk}{b\cdot d}=\frac{b\cdot d\cdot k^2}{b\cdot d}=k^2\left(2\right)\)

Từ (1) và (2) =>Đpcm

d)Đặt cũng như vậy 

Xét VT \(\frac{4\left(bk\right)^4+5b^4}{4\left(dk\right)^4+5d^4}=\frac{4b^4k^4+5b^4}{4d^4k^4+5d^4}=\frac{b^4\left(4k^4+5\right)}{d^4\left(4k+5\right)}=\frac{b^4}{d^4}\left(1\right)\)

Xét VP \(\frac{\left(bk\right)^2b^2}{\left(dk\right)^2d^2}=\frac{b^2k^2b^2}{d^2k^2d^2}=\frac{k^2b^4}{k^2d^4}=\frac{b^4}{d^4}\left(2\right)\)

Từ (1) và (2) =>Đpcm

a) \(\frac{a-b}{b}=\frac{c-d}{d}\)

Xét d. ( a - b ) = a . d - b . d

      b. ( c - d ) = b . c - b . d

Vì \(\frac{a}{b}=\frac{c}{d}\) => a . d = b . c

hay d. ( a - b ) = b. ( c - d )

=> \(\frac{a-b}{b}=\frac{c-d}{d}\)

Vậy \(\frac{a-b}{b}=\frac{c-d}{d}\)