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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{3a+5b}{3a-5b}=\frac{3bk+5b}{3bk-5b}=\frac{b\left(3k+5\right)}{b\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
\(\frac{3c+5d}{3c-5d}=\frac{3dk+5d}{3dk-5d}=\frac{d\left(3k+5\right)}{d\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
Vậy từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
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Ta có:
a/b=c/d => a/c=b/d=2a/2c=3b/3d
= 2a+3b/2c+3d=2a-3b/2c-3d
=> 2a+3b/2a-3b=2c+3d/2c-3d (ĐPCM)
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ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{d}{b}=\frac{c}{a}\Rightarrow\frac{c+d}{a+b}\Rightarrow\frac{3c+3d}{3a+3b}=\frac{3c-3d}{3a-3b}\)
\(\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)\(\left(điềuphảichứngminh\right)\)
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\(\frac{3a+5b}{2a-b}=\frac{3c+5d}{2c-d}\)
<=>\(\left(3a+5b\right)\left(2a-b\right)=\left(3c+5d\right)\left(2c-d\right)\)
<=>\(6ac+10ad-3bc-5bd=6ac+10bc-3ad-5bd\)
<=>\(10ad-3bc=10bc-3ad\)
<=>\(10ad-3bc-10bc+3ad=0\)
<=>\(13ad-13ac=0\)
<=>\(13ad=13ac\)
<=>\(ad=bc\)
<=>\(\frac{a}{b}=\frac{c}{d}\)(đpcm)
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Ta có: \(\frac{3a+5b}{2a-b}=\frac{3c+5d}{2c-d}\)
=> (3a+5b)(2c-d) =(2a-b)(3c+5d)
=> 3a(2c-d) +5b(2c-d) =2a(3c+5d) -b(3c+5d)
=> 6ac -3ad +10bc -5bd =6ac +10ad -3bc -5bd
=>7bc=7ad
=> bc=ad
=> a/b =c/d
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\frac{2bk+5b}{3bk-4b}=\frac{2dk+5d}{3dk-4d}\)
Xét VT \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\left(1\right)\)
Xét VP \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\left(2\right)\)
Từ (1) và (2) ta có Đpcm
\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
=> (3a + 5b)(3c - 5d) = (3a - 5b)(3c + 5d)
=> 9ac - 15ad + 15bc - 25bd = 9ac + 15ad - 15bc - 25bd
=> 9ac - 15ad + 15bc - 25bd - (9ac + 15ad - 15bc - 25bd) = 0
=> 9ac - 15ad + 15bc - 25bd - 9ac - 15ad + 15bc + 25bd = 0
=> (9ac - 9ac) + (-15ad - 15ad) + (15bc + 15bc) + (-25bd + 25bd) = 0
=> -30ad + 30bc = 0
=> -30ad = -30bc
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\)