Câu hỏi của Nguyễn Ngọc Sáng

Question

Cho :

$\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}$

CMR :

$\frac{a}{b}=\frac{c}{d}$

Anh Thư Đinh · Accepted Answer

ta có:

$\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}$

$\Rightarrow\frac{7a-11b}{7c-11d}=\frac{4a+5b}{4c+5d}$

$\Leftrightarrow\frac{7a}{7c}=\frac{11b}{11d}=\frac{4a}{4c}=\frac{5b}{5d}\Rightarrow\frac{a}{c}=\frac{b}{d}$

Mặt khác:

$\frac{a}{c}=\frac{b}{d}\Leftrightarrow\frac{a}{b}=\frac{c}{d}$

$\Rightarrowđpcm$Câu trả lời: ta có:

$\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}$

$\Rightarrow\frac{7a-11b}{7c-11d}=\frac{4a+5b}{4c+5d}$

$\Leftrightarrow\frac{7a}{7c}=\frac{11b}{11d}=\frac{4a}{4c}=\frac{5b}{5d}\Rightarrow\frac{a}{c}=\frac{b}{d}$

Mặt khác:

$\frac{a}{c}=\frac{b}{d}\Leftrightarrow\frac{a}{b}=\frac{c}{d}$

$\Rightarrowđpcm$

Mạnh Nguyễn · Answer

ta có:

7a−11b4a+5b=7c−11d4c+5d7a−11b4a+5b=7c−11d4c+5d

⇒7a−11b7c−11d=4a+5b4c+5d⇒7a−11b7c−11d=4a+5b4c+5d

⇔7a7c=11b11d=4a4c=5b5d⇒ac=bd⇔7a7c=11b11d=4a4c=5b5d⇒ac=bd

Mặt khác:

ac=bd⇔ab=cdac=bd⇔ab=cd

⇒đpcmCâu trả lời: ta có:

7a−11b4a+5b=7c−11d4c+5d7a−11b4a+5b=7c−11d4c+5d

⇒7a−11b7c−11d=4a+5b4c+5d⇒7a−11b7c−11d=4a+5b4c+5d

⇔7a7c=11b11d=4a4c=5b5d⇒ac=bd⇔7a7c=11b11d=4a4c=5b5d⇒ac=bd

Mặt khác:

ac=bd⇔ab=cdac=bd⇔ab=cd

⇒đpcm

Ôn tập toán 7