cc yêu cl

/b = c/d           => a/c = b/d 

=> a² / c² = b² / d²  = ab / cd

<=> 7a² / 7c² = 11a² / 11c²  = 8b² / 8d² = 3ab / 3cd

=> 7a² + 3ab / 7c² + 3cd = 11a² - 8b² / 11c² - 8d²

=> 7a² + 3ab / 11a² - 8b² = 7c² + 3cd / 11c² - 8d²              

=>  (đpcm)

cho \(\frac{a}{b}\)=\(\frac{c}{d}\)=k=> a=bk; c=dk

a. Vế trái =\(\frac{5a+3b}{5a-3b}\)=\(\frac{5bk+3b}{5bk-3b}\)=\(\frac{b\left(5k+3\right)}{b\left(5k-3\right)}\)=\(\frac{\left(5k+3\right)}{\left(5k-3\right)}\)(1)

Vế phải =\(\frac{5c+3d}{5c-3d}\)=\(\frac{5dk+3d}{5dk-3d}\)=\(\frac{d\left(5k+3\right)}{d\left(5k-3\right)}\)=\(\frac{\left(5k+3\right)}{\left(5k-3\right)}\)(2)

Từ (1) và (2) ta có\(\frac{5a+3b}{5a-3b}\)=\(\frac{5c+3d}{5c-3d}\)

b. Vế trái=\(\frac{7a^2+3ab}{11a^2-8b^2}\)=\(\frac{7b^2k^2+3b.k.b}{11b^2.k^2-8b^2}\)=\(\frac{b^2.k\left(7k+3\right)}{b^2\left(11k^2-8\right)}\)=\(\frac{k\left(7k+3\right)}{\left(11k^2-8\right)}\)(1)

Vế phải =\(\frac{7c^2+3cd}{11c^2-8d^2}\)=\(\frac{7d^2k^2+3d.k.d}{11d^2.k^2-8d^2}\)=\(\frac{d^2.k\left(7k+3\right)}{d^2\left(11k^2-8\right)}\)=\(\frac{k\left(7k+3\right)}{\left(11k^2-8\right)}\)(2)

Từ (1) và (2) ta có: \(\frac{7a^2+3ab}{11a^2-8b^2}\)=\(\frac{7c^2+3cd}{11c^2-8d^2}\)

giups mình với cảm ơn

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

VT : \(\frac{5a+3b}{5a-3b}\Rightarrow\frac{5kb+3b}{5ka-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\) (1)

VP : \(\frac{5c+3d}{5c-3d}=\frac{5kd+3d}{5kd-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\) (2)

Từ (1) và (2) => đpcm

#)Giải :

\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\Leftrightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\Leftrightarrow\frac{7a^2}{7c^2}=\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}\)

\(\Leftrightarrow\frac{7a^2+3ab}{7c^2+3cd}=\frac{11a^2-8b^2}{11a^2-8d^2}\Leftrightarrow\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)

#)Giải : (Cách 2)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7b^2k^2+3b^2k}{11b^2k^2-8d^2}=\frac{b^2\left(7k^2-3k\right)}{b^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\\\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\frac{d^2\left(7k^2-3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\end{cases}}}\)

=> đpcm

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{5a}{3b}=\dfrac{5c}{3d}\)

hay \(\dfrac{5a}{5c}=\dfrac{3b}{3d}\)

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

\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(\Leftrightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

hay \(\dfrac{5a+3n}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)(đpcm)