Question

Answer 1

54:

`a)`

`a/b=c/d => a/c=b/d`

Áp dụng t/c của DTCBN , ta đc ; 

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\\ \Rightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\\ \Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\left(đpcm\right)\)

`b)`

`a/b=c/d => a/c=b/d => a^2/c^2=b^2/d^2 `

Áp dụng tính chất của DTSBN , ta có :

\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\\ \Rightarrow\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\\ \Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)