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Bài `1:`
Số tự nhiên `a` chia cho `5` dư `4 ->` a = 5k + 4 (k ∈ N)`
Ta có: `a^2 = (5k + 4)^2`
` = 25k^2 + 40k + 16`
` = 25k^2 + 40k + 15 + 1`
` = 5.(5k^2 + 8k + 3) + 1`
Vì `5.(5k^2 + 8k + 3)` $\vdots$ `5`
`->` `5.(5k^2 + 8k + 3) + 1` chia `5` dư `1`
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Bài `2:`
`a) P = x^2 - 2x + 5`
` = x^2 - 2x + 1 + 4`
` = (x - 1)^2 + 4`
Vì `(x - 1)^2 ≥ 0 ∀ x`
`->` `(x - 1)^2 + 4 ≥ 4`
Dấu "=" xảy ra `↔` `x = 1`
Vậy `minP = 1` `↔` `x = 1`
`b) Q = 2x^2 – 6x = 2(x^2 – 3x) = 2(x^2 – 2. 3/2 x + 9/4 - 9/4) = 2[(x−3/2)^2 - 9/4 ] = 2(x−3/2)^2- 9/2`
Vì `(x−3/2)^2≥ 0` `->` `2(x−3/2)^2 ≥ 0 `
`->` `2(x−3/2)^2- 9/2 ≥ - 9/2`
Dấu "=" xảy ra `↔` `x = 3/2`
Vậy `minQ = -9/2` `↔` `x = 3/2`
`c) M = x^2 + y^2 – x + 6y + 10`
` = (x^2 -x + 1) + (y^2 + 6y + 9)`
` = (x - 1/2 )^2 + (y + 3)^2 + 3/4`
Vì `(x - 1/2)^2 ≥ 0 ∀ x`
`(y + 3)^2 ≥ 0 ∀ y`
`->` `(x - 1/2 )^2 + (y + 3)^2 + 3/4` `≥ 3/4`
Dấu "=" xảy ra `↔` `x = 1/2 ; y = -3 `
Vậy `minM = 3/4` `↔` `x = 1/2 ; y = -3 `
