Question

Giúp mik nhé mí bạn.

1) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . CM :

b) \(\dfrac{5a-3b}{3a+2b}=\dfrac{5c-3d}{3c+2d}\)

c) \(\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

d) \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)

e) \(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)

f) \(\dfrac{\left(a+c\right)^2}{a^2-c^2}=\dfrac{\left(b+d\right)^2}{b^2-d^2}\)

Làm được câu nào thì trả lời nhé . Thanks trước

Answer 1

Ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)

<=>\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{3a}{3c}=\dfrac{2b}{2d}\)

<=>\(\dfrac{5a-3b}{5c-3d}=\dfrac{3a-2b}{3c-2d}\)(đpcm)

Các câu sau tương tự

Answer 2

Nguyễn Thị Hồng Nhung chị làm bài f đc ko ạ ???

Answer 3

c/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
d/ tương tự câu b/
e/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2} \)(1)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)(đpcm)
f/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(1\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2-c^2}{b^2-d^2}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(a+c\right)^2}{a^2-c^2}=\dfrac{\left(b+d\right)^2}{b^2-d^2}\)