Đáp án:
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`c,`
Đặt `a/b = c/d = k`
`↔` \(\left\{ \begin{array}{l}\dfrac{a}{b}=k\\ \dfrac{c}{d}=k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}a=bk\\c=dk\end{array} \right.\)
Có : `(5a + 2b)/(5a - 2b)`
`= (5 . bk + 2b)/(5 . bk - 2b)`
`= (b (5k + 2) )/(b (5k - 2) )`
`= (5k + 2)/(5k - 2)` `(1)`
Có : `(5c + 2d)/(5c - 2d)`
`= (5 . dk + 2d)/(5 . dk - 2d)`
`= (d (5k + 2) )/(d (5k - 2) )`
`= (5k + 2)/(5k - 2)` `(2)`
Từ `(1), (2)`
`-> (5a + 2b)/(5a - 2b) = (5c + 2d)/(5c - 2d) (= (5k + 2)/(5k - 2) )`
`->` đpcm
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$\\$
`d,`
Đặt `a/b = c/d = k`
`↔` \(\left\{ \begin{array}{l}\dfrac{a}{b}=k\\ \dfrac{c}{d}=k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}a=bk\\c=dk\end{array} \right.\)
Có : `(a^2 + c^2)/(b^2 + d^2)`
`= ( (bk)^2 + (dk)^2)/(b^2 + d^2)`
`=(b^2 k^2 + d^2k^2)/(b^2 + d^2)`
`= (k^2 (b^2 + d^2) )/(b^2 + d^2)`
`= k^2` `(1)`
Có : `( [a + c]^2)/( [b + d]^2)`
`= ([ bk + dk]^2)/( [b + d]^2)`
`= ( [k (b + d)]^2)/( [b + d]^2)`
`= k^2/1^2`
`= k^2` `(2)`
Từ `(1), (2)`
`-> (a^2 + c^2)/(b^2 + d^2) = (a+c)^2/(b+d)^2 (= k^2)`
`->` đpcm
