Ta có :

\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\Leftrightarrow\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7a^2-5ac}{7b^2-5bd}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\\ Thaya=bk;c=dk,tacó:\)

\(\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7\cdot b^2\cdot k^2+5\cdot bk\cdot dk}{7b^2+5bd}=\dfrac{k^2\cdot\left(7b^2+5ac\right)}{7b^2+5ac}=k^2\left(1\right)\)

\(\dfrac{7a^2-5ac}{7b^2-5bd}=\dfrac{7\cdot b^2\cdot k^2-5\cdot bk\cdot dk}{7b^2-5bd}=\dfrac{k^2\cdot\left(7b^2-5ac\right)}{7b^2-5ac}=k^2\left(2\right)\)

từ (1) và (2) \(\RightarrowĐpcm\)

vì sao phải đổi mẫu tử của 2 phs kia ?

Với \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{b}.\)\(\dfrac{c}{d}=\dfrac{ac}{bd}=\dfrac{aa}{bb}=\dfrac{a^2}{b^2}\)
Ta có : \(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\)

=> \(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}=\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7a^2-5ac}{7b^2-5bd}\) (1)

Từ (1) => \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2-5bd}{7b^2-5bd}\) (ĐPCM)

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

Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(VT=\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{a\left(7a+5c\right)}{a\left(7a-5c\right)}=\dfrac{7ck+5c}{7ck-5c}=\dfrac{c\left(7k+5\right)}{c\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(1\right)\)

\(VP=\dfrac{7b^2+5bd}{7b^2-5bd}=\dfrac{b\left(7b+5d\right)}{b\left(7b-5d\right)}=\dfrac{7dk+5d}{7dk-5d}=\dfrac{d\left(7k+5\right)}{d\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)

\(\Rightarrow\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\left(đpcm\right)\)

Vậy \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\)

Bài 2: 

Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)

\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)

a) Ta co: a/b = c/d= k

=> a=bk

c=dk

Ta co: a-b/a+b = bk-b/bk+b = b(k-1)/b(k+1) = k-1/k+1 (1)

Ta co: c-d/c+d = dk-d/dk+d = d(k-1)/d(k+1) = k-1/k+1 (2)

Tu (1) va (2)

=> a-b/a+b=c-d/c+d

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

a) Từ (*) ta có:

\(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\) (1)

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

Từ (1) và (2) suy ra \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

b) Từ (*) ta có:

\(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{b\left(7k-4\right)}{b\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (3)

\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{d\left(7k-4\right)}{d\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (4)

Từ (3) và (4) suy ra \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)

c) Từ (*) ta có:

\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (5)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (6)

\(\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}=\dfrac{\left[\left(dk\right)-\left(bk\right)\right]^2}{\left(d-b\right)^2}=\dfrac{\left[k\left(d-b\right)\right]^2}{\left(d-b\right)^2}=k^2\) (7)

Từ (5), (6) và (7) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{3a^2+5ac}{3a^2-5ac}=\dfrac{3b^2k^2+5\cdot bk\cdot dk}{3b^2k^2-5\cdot bk\cdot dk}=\dfrac{3b^2k^2+5bdk^2}{3b^2k^2-5bdk^2}=\dfrac{3b^2+5bd}{3b^2-5bd}\)

từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

ta có:\(\dfrac{5a+3b}{7a-2b}=\dfrac{5.ck+3.dk}{7.ck-2.dk}=\dfrac{k.\left(5c+3d\right)}{k.\left(7c-2d\right)}=\dfrac{5c+3d}{7c-2d}\)Vậy \(\dfrac{5a+3b}{7a-2b}=\dfrac{5c+3d}{7c-2d}\left(đpcm\right)\)

b) từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

ta có:\(\dfrac{7a^2+3ab}{11a^2+8.b^2}=\dfrac{7.c^2.k^2+3.c.d.k^2}{11.c^2.k^2+8.d^2.k^2}=\dfrac{k^2.\left(7.c^2+3.c.d\right)}{k^{2.}\left(11.c^2+8.d^2\right)}\) vậy .......

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

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

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

Từ (1).(2)=>......

Nhấn vào đây

Ta có:

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

a) \(\dfrac{3a+5c}{3b+5d}=\dfrac{3\cdot bk+5\cdot dk}{3b+5d}=\dfrac{k\left(3b+5d\right)}{3b+5d}=k\) (1)

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

Từ (1) và (2) \(\Rightarrow\dfrac{3a+5c}{3b+5d}=\dfrac{a-2c}{b-2d}\left(dpcm\right)\)

b) \(\dfrac{a^2-b^2}{ab}=\dfrac{\left(bk\right)^2-b^2}{bk\cdot b}=\dfrac{b^2k^2-b^2}{b^2k}=\dfrac{b^2\left(k-1\right)}{b^2k}=\dfrac{k-1}{k}\)(1)

\(\dfrac{c^2-d^2}{cd}=\dfrac{\left(dk\right)^2-d^2}{dk\cdot d}=\dfrac{d^2k^2-d^2}{d^2k}=\dfrac{d^2\left(k-1\right)}{d^2k}=\dfrac{k-1}{k}\) (2)

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

c) \(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\dfrac{b^3}{d^3}\) (1)

\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\) (2)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\left(dpcm\right)\)

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