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Bài `1.`
Vì số tự nhiên `a` chia `5` dư `4`
`-> a=5k + 4 (k ∈ NN)`
`->a^2 = (5k+4)^2`
`->a^2 =25k^2 + 40k + 16`
`->a^2 = 25k^2 + 40 k+15 + 1 = 5 (5k^2 + 8k + 3) + 1`
`->a^2` chia `5` dư `1`
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Bài `2.`
`a,`
`P=x^2 -2x+5`
`->P=x^2 - 2.x.1+1^2+4`
`->P=(x-1)^2+4≥4∀x`
Dấu "`=`" xảy ra khi :
`(x-1)^2=0 ↔ x-1=0 ↔x=1`
Vậy `min P=4↔x=1`
`b,`
`Q=2x^2 -6x`
`->Q=2(x^2 - 3x)`
`->Q=2 (x^2 - 2 . x . 3/2 + 9/4 -9/4)`
`->Q=2 (x-3/2)^2 - 9/2 ≥ (-9)/2∀x`
Dấu "`=`" xảy ra khi :
`(x-3/2)^2=0 ↔ x-3/2=0 ↔x=3/2`
Vậy `min Q=(-9)/2 ↔x=3/2`
`c,`
`M=x^2 +y^2 - x + 6y+10`
`->M=(x^2 - x + 1/4) + (y^2 + 6y + 9) +3/4`
`->Q=(x-1/2)^2 + (y+3)^2 + 3/4 ≥ 3/4 ∀x,y`
Dấu "`=`" xảy ra khi :
`(x-1/2)^2=0 , (y+3)^2=0`
`↔ x-1/2=0,y+3=0`
`↔x=1/2,y=-3`
Vậy `min Q=3/4 ↔x=1/2,y=-3`
