Đáp án:
`a,`
`(ac)/(bd) = (a^2 + c^2)/(b^2 + d^2)`
`b,`
`(a^2 + 5c^2 - 3ac)/(b^2 + 5d^2 - 3bd) = (a^2 + ac)/(b^2 + bd)`
Giải thích các bước giải:
`a,`
Đặt `a/c=b/d=k`
$\bullet$ `a/c=k ↔ a = ck`
$\bullet$ `b/d=k ↔ b = dk`
Có : `(ac)/(bd)`
`= (ck . c)/(dk . d)`
`= (c^2k)/(d^2k)`
`= c^2/d^2` `(1)`
Có : `(a^2 + c^2)/(b^2+d^2)`
`= ( (ck)^2 + c^2)/( (dk)^2 + d^2)`
`= (c^2k^2 + c^2)/(d^2k^2+d^2)`
`= (c^2 (k^2 + 1) )/(d^2 (k^2+1) )`
`= c^2/d^2` `(2)`
Từ `(1), (2)`
`-> (ac)/(bd) = (a^2 + c^2)/(b^2 + d^2) (=c^2/d^2)`
`->` đpcm
`b,`
Đặt `a/b=b/d=k`
$\bullet$ `a/c = k ↔ a = ck`
$\bullet$ `b/d=k ↔ b = dk`
Có : `(a^2 + 5c^2 - 3ac)/(b^2 + 5d^2- 3bd)`
`= ( (ck)^2 + 5c^2 - 3 . ck . c)/( (dk)^2 + 5d^2 - 3 . dk . d)`
`= (c^2k^2 + 5c^2 - 3c^2k)/(d^2k^2 + 5d^2 - 3d^2k)`
`= (c^2 (k^2 + 5 - 3k) )/(d^2 (k^2 + 5 - 3k) )`
`= c^2/d^2` `(1)`
Có : `(a^2 + ac)/(b^2 + bd)`
`= ( (ck^2) + ck . c)/( (dk)^2 + dk . d)`
`= (c^2k^2 + c^2k)/(d^2k^2 + d^2k)`
`= (c^2 (k^2+k) )/(d^2 (k^2+k) )`
`= c^2/d^2` `(2)`
Từ `(1), (2)`
`-> (a^2 + 5c^2 - 3ac)/(b^2 + 5d^2 - 3bd) = (a^2 + ac)/(b^2 + bd) (=c^2/d^2)`
`->` đpcm
