Câu hỏi của Quỳnh Mai Đỗ

Question

Cho tỉ lệ thức$\dfrac{a}{b}=\dfrac{c}{d}$ .Chứng minh;

$\dfrac{a^2}{b^2}=\dfrac{a^2-ac}{b^2-bd}
$

Lightning Farron · Accepted Answer

Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk$

Xét $VT=\dfrac{a^2}{b^2}=\dfrac{\left(bk\right)^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\left(1\right)$

Xét $VP=\dfrac{a^2-ac}{b^2-bd}=\dfrac{\left(bk\right)^2-bk\cdot dk}{b^2-bd}=\dfrac{b^2k^2-bdk^2}{b^2-bd}$

$=\dfrac{k^2\left(b^2-bd\right)}{b^2-bd}=k^2\left(2\right)$

Từ $\left(1\right);\left(2\right)$ ta có ĐPCMCâu trả lời: Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk$

Xét $VT=\dfrac{a^2}{b^2}=\dfrac{\left(bk\right)^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\left(1\right)$

Xét $VP=\dfrac{a^2-ac}{b^2-bd}=\dfrac{\left(bk\right)^2-bk\cdot dk}{b^2-bd}=\dfrac{b^2k^2-bdk^2}{b^2-bd}$

$=\dfrac{k^2\left(b^2-bd\right)}{b^2-bd}=k^2\left(2\right)$

Từ $\left(1\right);\left(2\right)$ ta có ĐPCM