Bài1:CMR từ tỉ lệ thức a/b=c/d suy ra tỉ lệ thức 5a+4b/5a-4b=5c+4d/5c-4d Bài 2: a)CMR nếu a/b=c/d thì a^2+b^2/b^2+c^2=a/c b)Nếu a/b=b/c=c/d thì(a+b-c/b+c-d)^3=a/d

baytayson

2 năm trước

Bài1:CMR từ tỉ lệ thức a/b=c/d suy ra tỉ lệ thức 5a+4b/5a-4b=5c+4d/5c-4d Bài 2: a)CMR nếu a/b=c/d thì a^2+b^2/b^2+c^2=a/c b)Nếu a/b=b/c=c/d thì(a+b-c/b+c-d)^3=a/d

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luongngocanh

`Bài 1:`

`Có`

`a/b = c/d = k`

`→ a = bk; c = dk`

`→ (5a + 4b)/(5a - 4b) = (5bk + 4b)/(5bk - 4b) = (b(5k + 4))/(b(5k - 4)) = (5k + 4)/(5k - 4) `

`(5c + 4d)/(5c - 4d) = (5dk + 4d)/(5dk - 4d) = (d(5k + 4))/(d(5k - 4)) = (5k + 4)/(5k - 4) `

`⇒ (5a + 4b)/(5a - 4b) = (5c + 4d)/(5c - 4d)`

`Bài 2:`

`a)` Đề sai

`b) a/b = b/c = c/d = (a + b - c)/(b + c - d) `

`→ (a/b)^3 = ((a + b - c)/(b + c - d))^3`

`→ a/d = ((a + b - c)/(b + c - d))^3 (Đpcm)`

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didi2525

Đáp án:

Giải thích các bước giải:

Bài 1:

Đặt `a/b=c/d=k`

`=> a=bk, c=dk`

`=>`` (5a+4b)/(5a-4b)= (5bk+4b)/(5bk-4b)=(b(5k+4))/(b(5k-4))=(5k+4)/(5k-4)` `(**)`

`=>`` (5c+4d)/(5c-4d)= (5dk+4d)/(5dk-4d)=(d(5k+4))/(d(5k-4))=(5k+4)/(5k-4)` `(***)`

Từ `(**)` và `(***)` `=>` `(5a+4b)/(5a-4b)=(5c+4d)/(5c-4d)` `(đpcm)`

Bài 2:

`a)`

`b)`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`a/b=b/c=c/d=(a+b+c)/(b+c+d)`

`=> (a/b)^3=((a+b+c)/(b+c+d))^3`

Vì `(a/b)^3=a/d` (Rút gọn)

`=>` `((a+b-c)/(b+c-d))^3=a/d`

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