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\(\left\{{}\begin{matrix}a^2+ab+\dfrac{b^2}{3}=25\\c^2+\dfrac{b^2}{3}=9\\a^2+ac+c^2=16\end{matrix}\right.\)
\(\Rightarrow a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2\)
\(\Rightarrow ab=2c^2+ac\)
Biến đổi 1 chút là ra
\(\rightarrowđpcm\)
\(\hept{\begin{cases}a^2+ab+\frac{b^2}{3}=25\\c^2+\frac{b^2}{3}=9\end{cases}}\Rightarrow a^2+ac-c^2=16\)
\(\Rightarrow a^2+ab-c^2=a^2+ac+c^2\left(=16\right)\)
\(\Rightarrow ab-c^2=ac+c^2\)
\(\Rightarrow ab=ac+2c^2\)
\(\Rightarrow ab+ac=2ac+2c^2\)
\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\)
\(\Leftrightarrow\frac{2c}{a}=\frac{b+c}{a+c}\left(đpcm\right)\)
Có \(a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\left(=25\right)\)
\(\Rightarrow a^2+ab+\frac{b^2}{3}=2c^2+\frac{b^2}{3}+a^2+ac\\ \Rightarrow ab=2c^2+ac\\ \Rightarrow ab+ac=2c^2+2ac\\ \Rightarrow a\left(b+c\right)=2c\left(a+c\right)\\ \Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\)
Ta có:
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
Ta có : \(b^2=ac\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}\) (1)
\(c^2=bd\)
\(\Rightarrow\frac{b}{c}=\frac{c}{d}\) (2)
Từ (1) và (2) suy ra : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\) , \(\frac{b}{c}.\frac{b}{c}.\frac{b}{c}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\) và \(\frac{c}{d}.\frac{c}{d}.\frac{c}{d}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{a}{d}\) , \(\frac{b^3}{c^3}=\frac{a}{d}\) và \(\frac{c^3}{d^3}=\frac{a}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Vậy \(\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)