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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b\left(2k+13\right)}{b\left(3k-7\right)}=\frac{2k+13}{3k-7}\left(1\right)\)
\(\frac{2c+13d}{3c-7d}=\frac{2dk+13d}{3dk-7d}=\frac{d\left(2k+13\right)}{d\left(3k-7\right)}=\frac{2k+13}{3k-7}\left(2\right)\)
Từ \(\left(1\right)\) và (2) \(\Rightarrow\frac{a}{b}=\frac{c}{d}\)( đpcm )
Chúc bạn học tốt !!!
Từ \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)\(\Rightarrow\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}=\frac{2a}{2c}=\frac{13b}{13d}=\frac{3a}{3c}=\frac{7b}{7d}=\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Suy ra : \(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b.\left(2k+13\right)}{b.\left(3k-7\right)}=\frac{2k+13}{3k-7}\)
\(\frac{2c+13d}{3c-7d}=\frac{2dk+13d}{3dk-7d}=\frac{d\left(2k+13\right)}{d\left(3k-7\right)}=\frac{2k+13}{3k-7}\)
Vậy \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\) Khi : \(\frac{a}{b}=\frac{c}{d}\)
ta có : \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)
<=> (2a+13b)(3c-7d)=(2c+13d)(7a-7b)
<=>6ac-14ad+39bc-91bd=6c-14bc+39ab-91bd
<=>39bc-14ab=39ab-14bc
<=> bc=ab
<=>\(\frac{a}{b}=\frac{c}{d}\)
Ta có: \(\frac{2a+13b}{3a-7c}=\frac{2c+13d}{3a-7d}\)
\(\Rightarrow\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}=\frac{2a+13b+3a-7b}{2c+13d+3c-7d}=\frac{5a+6b}{5c+6d}\)
\(\Rightarrow\frac{5a+6b}{5c+6d}\Rightarrow\frac{5a}{5c}=\frac{6b}{6d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\left(đpcm\right)\)
\(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\Leftrightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(3a-7b\right)\left(2c+13d\right)\)
\(\Leftrightarrow6ac-14ad+39bc-91bd=6ac+39ad-14bc-91bd\)
\(\Leftrightarrow-14ad+39bc=39ad-14bc\)\(\Leftrightarrow53bc=53ad\)\(\Leftrightarrow bc=ad\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
ĐẶT \(\frac{a}{b}\)= \(\frac{c}{d}\)là k
suy ra a=kb; c=kd
ta có:\(\frac{2a+13b}{3a-7b}\)= \(\frac{2kb+13b}{3kb-7b}\)= \(\frac{b\left(2k+13\right)}{b\left(3k-7b\right)}\)=\(\frac{2k+13}{3k-7}\) (1)
\(\frac{2c+13d}{3c-7d}\)=\(\frac{2kd+13d}{3kd-7d}\)=\(\frac{d\left(2k+13\right)}{d\left(3k-7\right)}\)=\(\frac{2k+13}{3k-7}\) (2)
từ (1) và (2) suy ra \(\frac{2a+13b}{3a-17b}\)=\(\frac{2c+13d}{3c-7d}\)