Câu hỏi của Bùi Hiền Thảo

Question

Biết: a2+ab+$\dfrac{b^2}{3}$=25; c2+$\dfrac{b^2}{3}$=9; a2+ac+c2=16 và $a\ne0,c\ne0,a\ne-c$

Chứng minh rằng: $\dfrac{2c}{a}=\dfrac{b+c}{a+c}$

Mashiro Shiina · Accepted Answer

Ta có:

$a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2$

$\Rightarrow a^2+ab+\dfrac{b^2}{3}=2c^2+\dfrac{b^2}{3}+a^2+ac$

$\Rightarrow ab=2c^2+ac$

$\Rightarrow ab+ac=2ac+2c^2$

$\Rightarrow a\left(b+c\right)=2c\left(a+c\right)$

$\Rightarrow\dfrac{2c}{a}=\dfrac{b+c}{a+c}\left(đpcm\right)$Câu trả lời: Ta có:

$a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2$

$\Rightarrow a^2+ab+\dfrac{b^2}{3}=2c^2+\dfrac{b^2}{3}+a^2+ac$

$\Rightarrow ab=2c^2+ac$

$\Rightarrow ab+ac=2ac+2c^2$

$\Rightarrow a\left(b+c\right)=2c\left(a+c\right)$

$\Rightarrow\dfrac{2c}{a}=\dfrac{b+c}{a+c}\left(đpcm\right)$

Đại số lớp 7