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Bài `1.`
Đặt `a/b=c/d=k (k \ne 0)`
`->a/b=k->a=bk`
và `c/d=k->c=dk`
`a,`
`(2a - 3b)/(2a + 3b) = (2bk - 3b)/(2bk + 3b) = (b (2k-3) )/(b (2k+3) ) = (2k-3)/(2k+3)` (1)
`(2c-3d)/(2c+3d)=(2dk - 3d)/(2dk + 3d) = (d (2k-3) )/(d (2k+3) )=(2k-3)/(2k+3)` (2)
Từ (1), (2) `-> (2a-3b)/(2a+3b)=(2c-3d)/(2c+3d) (=(2k-3)/(2k+3) )`
`b,`
`( (a-b)/(c-d) )^4=( (bk-b)/(dk-d) )^4 = ( (b (k-1) )/(d (k-1) ) )^4 = (b/d)^4=b^4/d^4` (1)
`(a^4 + b^4)/(c^4 +d^4) = (b^2 k^4 + b^4)/(d^4 k^4 +d^4) = (b^4 (k^4+1) )/(d^4 (k^4+1) ) = b^4/d^4` (2)
Từ `(1), (2) -> ( (a-b)/(c-d) )^4=(a^4 + b^4)/(c^4+d^4) (=b^4/d^4)`
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Bài `2.`
Vận dụng : `(a+b)^2 =a^2 +2ab + b^2`
`(a-b)^2 = a^2 - 2ab + b^2`
Chứng minh : `(a+b)^2=a^2 +2ab + b^2`
Xét VT : `(a+b)^2 = (a+b)(a+b)=a(a+b)+b (a+b)=a^2 +ab+ab+b^2=a^2 + 2ab+b^2` (Bằng VP)
`-> (a+b)^2=a^2 +2ab + b^2`
Chứng minh :`(a-b)^2 = a^2 - 2ab + b^2`
Xét VT : `(a-b)^2=(a-b)(a-b)=a(a-b)-b(a-b)=a^2 - ab - ab + b^2=a^2 - 2ab + b^2` (Bằng VP)
`-> (a-b)^2=a^2 - 2ab + b^2`
`a,`
`34^2 + 66^2 + 66 . 68`
`= 34^2 + 66 . 68 + 66^2`
`= 34^2 + 2 . 34 . 66 + 66^2`
`= (34+66)^2`
`= 100^2`
`= 10000`
`b,`
`74^2 + 24^2 - 48 . 74`
`= 74^2 - 48. 74 + 24^2`
`= 74^2 - 2 . 24 . 74 + 24^2`
`= (74 - 24)^2`
`= 50^2`
`= 2500`
