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Đặt `a/b=c/d=k(k\ne 0)`
`->a=bk,c=dk`
`a,`
`(5a+3b)/(5a-3b)=(5bk+3b)/(5bk-3b)=(b(5k+3))/(b(5k-3))=(5k+3)/(5k-3)`
`(5c+3d)/(5c-3d)=(5dk+3d)/(5dk-3d)=(d(5k+3))/(d(5k-3))=(5k+3)/(5k-3)`
`-> (5a+3b)/(5a-3b)=(5c+3d)/(5c-3d)(=(5k+3)/(5k-3))`
`b,`
`(7a^2+3ab)/(11a^2 - 8b^2)`
`= (7b^2k^2 + 3b^2k)/(11b^2k^2 - 8b^2)`
`= (b^2 (7k^2 + 3k))/(b^2 (11k^2 - 8))`
`= (7k^2 +3k)/(11k^2-8)` (1)
`(7c^2+3cd)/(11c^2 - 8d^2)`
`=(7d^2k^2 +3d^2k)/(11d^2k^2 - 8d^2)`
`= (d^2 (7k^2 +3k))/(d^2(11k^2 - 8))`
`= (7k^2+3k)/(11k^2 - 8)` (2)
Từ (1)(2)
`-> (7a^2+3ab)/(11a^2-8b^2)=(7c^2+3cd)/(11c^2-8d^2)(=(7k^2+3k)/(11k^2-8))`
