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\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\\ \Leftrightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc=ad-bd=bc-bd=d.\left(a-b\right)=b.\left(c-d\right)\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)Đúng 100% tick nha
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Rightarrow\frac{q^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)
=> \(\frac{a^2}{4}=4\Rightarrow a^2=4.4=16\Rightarrow a=+-4\)
=>\(\frac{b^2}{9}=4\Rightarrow b^2=4.9=36\Rightarrow b=+-6\)
=>\(\frac{2c^2}{32}=4\Rightarrow c^2=4.32:2=64\Rightarrow c=+-8\)
Câu 2 :
Ta có : \(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
Ta có:
\(\frac{a+b}{a-b}=\frac{c+a}{c-a}\Rightarrow\left(a+b\right)\left(c-a\right)=\left(c+a\right)\left(a-b\right)\)
\(\Rightarrow ac+bc-a^2-ba=ca+a^2-bc-ba\)
\(\Rightarrow2a^2=2bc\)
\(\Rightarrow a^2=bc\)
\(\Rightarrowđpcm\)
có \(\frac{a}{b}=\frac{c}{d}\)
\(=\orbr{\begin{cases}\frac{a+c}{b+d}\\\frac{a-c}{b-d}\end{cases}}\)
\(\Rightarrow\frac{a+c}{b+d}=\frac{a-c}{b-d}\left(=\orbr{\begin{cases}\frac{a}{b}\\\frac{c}{d}\end{cases}}\right)\)