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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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Có \(\frac{a}{b}=\frac{c}{d}\)
=> \(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}\)
Áp dụng dãy tỉ số bằng nhau
=> \(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{3a+b}{3c+d}\)
=> \(\frac{a}{c}=\frac{3a+b}{3c+d}\)
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
=> Đpcm
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=b\times k\) ; \(c=d\times k\)
Ta có :
\(\frac{a}{3a+b}=\frac{b\times k}{3\times b\times k+b}=\frac{b\times k}{b\left(3k+1\right)}=\frac{k}{3k+1}\) (1)
\(\frac{c}{3c+d}=\frac{d\times k}{3\times d\times k+d}=\frac{d\times k}{d\left(3k+1\right)}=\frac{k}{3k+1}\) (1)
Từ (1) và (2) \(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
k mk nha bạn !
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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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Do \(\frac{a}{b}=\frac{c}{d}\) => \(\frac{b}{a}=\frac{d}{c}\)
=> \(3+\frac{b}{a}=3+\frac{d}{c}\)
=> \(\frac{3a+b}{a}=\frac{3c+d}{c}\)
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3b+3c+3d+3a}=\frac{a+b+c+d}{3\cdot\left(b+c+d+a\right)}=\frac{1}{3}\)
Do đó :
\(\frac{a}{3b}=\frac{1}{3}\Rightarrow\frac{a}{b}.\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{a}{b}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow a=b\)
\(\frac{b}{3c}=\frac{1}{3}\Rightarrow\frac{b}{c}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{b}{c}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow b=c\)
\(\frac{c}{3d}=\frac{1}{3}\Rightarrow\frac{c}{d}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{c}{d}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow c=d\)
\(\frac{d}{3a}=\frac{1}{3}\Rightarrow\frac{d}{a}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{d}{a}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow d=a\)
\(\Rightarrow a=b=c=d\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow a=bk,c=dk\)
\(\frac{3a+7b}{3c-7d}=\frac{3bk+7b}{3dk+7d}=\frac{b\left(3k+7\right)}{d\left(3k+7\right)}=\frac{b}{d}\)(1)
\(\frac{3a-7b}{3c-7d}=\frac{3bk-7b}{3dk-7d}=\frac{b\left(3k-7\right)}{d\left(3k-7\right)}=\frac{b}{d}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{3a+7b}{3c+7d}=\frac{3a-7b}{3c-7d}\)