Toán Lớp 9: a+b+c=6
0 ≤a;b;c ≤4
tìm maxP=a^2+b^2+c^2+ab+ac+bc
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Toán Lớp 9: a+b+c=6
0 ≤a;b;c ≤4
tìm maxP=a^2+b^2+c^2+ab+ac+bc
Comments ( 2 )
Ta có:
P = a^2 + b^2 + c^2 + ab + ac + bc
= (a + b + c)^2 – 2ab – 2bc – 2ac + ab + ac + bc
= 6^2 – ab – ac – bc
= 36 – (ab + ac + bc)
Vì: a; b; c ≤ 4
⇒ (a – 4)(b – 4)(c – 4) ≤ 0
⇔ abc – 4ac – 4bc – 64 – 4ab + 16a + 16b + 16c ≤ 0
⇔ abc – 4(ab + ac + bc) + 16(a + b + c) – 64 ≤ 0
Vì: 0 ≤ a; b; c
⇒ abc ≥ 0
⇔ – abc ≤ 0
Cộng 2 BĐT cùng chiều theo từng vế, ta được:
abc – 4(ab + ac + bc) + 16(a + b + c) – 64 – abc ≤ 0
⇔ – 4(ab + ac + bc) + 16(a + b + c) ≤ 64
⇔ – 4(ab + ac + bc) + 16.6 ≤ 64
⇔ – 4(ab + ac + bc) ≤ – 32
⇔ – (ab + ac + bc) ≤ – 8
⇒ 36 – (ab + ac + bc) ≤ 36 – 8 = 28
Vậy \text{max}_\text{p} là 28 khi: (a; b; c) = (0; 2; 4) và các hoán vị
Ta có:
P=a2+b2+c2+ab+ac+bcP=a2+b2+c2+ab+ac+bc
=(a+b+c)2−2ab−2bc−2ac+ab+ac+bc=(a+b+c)2-2ab-2bc-2ac+ab+ac+bc
=62−ab−ac−bc=62-ab-ac-bc
=36−(ab+ac+bc)=36-(ab+ac+bc)
Vì: a;b;c≤4a;b;c≤4
⇒(a−4)(b−4)(c−4)≤0⇒(a-4)(b-4)(c-4)≤0
⇔abc−4ac−4bc−64−4ab+16a+16b+16c≤0⇔abc-4ac-4bc-64-4ab+16a+16b+16c≤0
⇔abc−4(ab+ac+bc)+16(a+b+c)−64≤0⇔abc-4(ab+ac+bc)+16(a+b+c)-64≤0
Vì: 0≤a;b;c0≤a;b;c
⇒abc≥0⇒abc≥0
⇔−abc≤0⇔-abc≤0
Cộng 22 BĐT cùng chiều theo từng vế, ta được:
abc−4(ab+ac+bc)+16(a+b+c)−64−abc≤0abc-4(ab+ac+bc)+16(a+b+c)-64-abc≤0
⇔−4(ab+ac+bc)+16(a+b+c)≤64⇔-4(ab+ac+bc)+16(a+b+c)≤64
⇔−4(ab+ac+bc)+16.6≤64⇔-4(ab+ac+bc)+16.6≤64
⇔−4(ab+ac+bc)≤−32⇔-4(ab+ac+bc)≤-32
⇔−(ab+ac+bc)≤−8⇔-(ab+ac+bc)≤-8
⇒36−(ab+ac+bc)≤36−8=28⇒36-(ab+ac+bc)≤36-8=28
Vậy maxpmaxp là 2828 khi: (a;b;c)=(0;2;4)(a;b;c)=(0;2;4) và các hoán vị