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Đặt `a/b=c/d=k (k \ne 0)`
`->a/b=k ->a=bk`
và `c/d=k ->c=dk`
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`a,`
Có : `(4a+3c)/(4b+3d)`
`= (4bk + 3dk)/(4b + 3d)`
`= (k (4b+3d) )/(4b+3d)`
`= k` (1)
Có : `(4a-3c)/(4b-3d)`
`= (4bk - 3dk)/(4b-3d)`
`= (k (4b-3d) )/(4b-3d)`
`= k` (2)
Từ (1), (2)
`-> (4a+3c)/(4b+3d) = (4a-3c)/(4b-3d) (=k)`
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`b,`
Có : `(a+c)^2/(b+d)^2`
`= (bk +dk)^2/(b+d)^2`
`= [k (b+d)]^2/(b+d)^2`
`= (k^2 (b+d)^2 )/(b+d)^2`
`= k^2` (1)
Có : `(a^2 - c^2)/(b^2 - d^2)`
`= (b^2k^2 - d^2k^2)/(b^2 - d^2)`
`= (k^2 (b^2 - d^2) )/(b^2 - d^2)`
`= k^2` (2)
Từ (1), (2)
`-> (a+c)^2/(b+d)^2 = (a^2 -c^2)/(b^2- d^2) (=k^2)`
