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Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{5a}{5c}=\frac{2b}{2d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a+2b}{5c+2d}=\frac{5a-2b}{5c-2d}\)
\(\Rightarrow\frac{5a+2b}{5a-2b}=\frac{5c+2d}{5c-2d}\left(đpcm\right)\)
ta có:
\(\frac{5a+2b}{5a-2b}=\frac{5c+2d}{5c-2d}\Rightarrow\frac{5a+2b}{5c+2d}=\frac{5a-2b}{5c-2d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a-2b}{5c-2d}=\frac{5a+2b}{5c+2d}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^2b}{c^2d}=\frac{2b^3}{2d^3}=\frac{a^3+2b^3}{c^3+2d^3}\)
=>đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : ab=cd⇔ad=bc⇔4ad=4bc⇔2ad+2ad=2bc+2bcab=cd⇔ad=bc⇔4ad=4bc⇔2ad+2ad=2bc+2bc
⇔2ad−2bc=2bc−2ad⇔ac+2ad−2bc−4bd=ac+2bc−2ad−4bd⇔2ad−2bc=2bc−2ad⇔ac+2ad−2bc−4bd=ac+2bc−2ad−4bd
⇔(c+2d)(a−2b)=(a+2b)(c−2d)⇔a+2bc+2d=a−2bc−2d(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
a/b =c/d
⟹a/c=b/d
Áp dụng tính chất dãy tỉ số bằng nhau ta có
a/c=b/d=5a+2b/5c+2b=5a-2b/5c-2d
Vì 5a=2b/5c=2d=5a-2b/5c-2d
⟹5a+2b/5a-2b=5c+2d/5c-2d
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}.\)
\(\Rightarrow\frac{5a}{5c}=\frac{2b}{2d}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a+2b}{5c+2d}\) (1)
\(\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a-2b}{5c-2d}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{5a+2b}{5c+2d}=\frac{5a-2b}{5c-2d}\)
\(\Rightarrow\frac{5a+2b}{5a-2b}=\frac{5c+2d}{5c-2d}\left(đpcm\right).\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này có 2 cách nè:
Cách 1:
\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}=\frac{\left(a+b\right)-\left(a-2b\right)}{\left(c+d\right)-\left(c-2d\right)}=\frac{a+b-a+2b}{c+d-c+2d}=\frac{3b}{3d}=\frac{b}{d}\left(1\right)\)
\(\frac{a+b}{c+a}=\frac{a-2b}{c-d}=\frac{2a+2b}{2c+2d}=\frac{2-ab+2a+2b}{c-2d+2c+2d}=\frac{3a}{3c}=\frac{a}{c}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\frac{b}{d}=\frac{a}{c}\) Suy ra: \(\frac{a}{b}=\frac{c}{d}\)
Cách 2:
\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
\(\Rightarrow\left(a+b\right).\left(c-2d\right)=\left(a-2b\right).\left(c+d\right)\)
\(\Rightarrow a.c-2a.d+b.c-2b.d=a.c+a.d-2b.c-2b.d\)
\(\Rightarrow a.c-a.c-2a.d-a.d+b.c-2b.c-2b.d+2b.d=0\)
\(\Rightarrow-3a.d+3b.d=0\)
\(\Rightarrow3b.c=3a.d\)
\(\Rightarrow b.c=a.d\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\Leftrightarrow\frac{a+b}{a}=\frac{c+d}{c}\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
từ \(\frac{a+b}{a}=\frac{c+d}{c}\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
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