cho x,y,z tm x^2 + y^2 + z^2 – 2x-4y-6z nho hon hoac bang -11 cm 3 nho hon hay bang x+y+z nho hon bang 9 07/11/2021 Bởi Rose cho x,y,z tm x^2 + y^2 + z^2 – 2x-4y-6z nho hon hoac bang -11 cm 3 nho hon hay bang x+y+z nho hon bang 9
x² + y² + z² – 2x – 4y -6z ≤ -11 ⇔ x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 ≤ 3 ⇔ (x – 1)² + (y – 2)² + (z – 3)² ≤ 3 Mà [(x – 1) + (y – 2) + (z – 3)]² ≤ 3[(x – 1)² + (y – 2)² + (z – 3)²] = 9 ⇒ – 3 ≤ (x – 1) + (y – 2) + (z – 3) ≤ 3 ⇔ 3 ≤ x + y + z ≤ 9 Bình luận
x² + y² + z² – 2x – 4y -6z ≤ -11 <=> x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 ≤ 3 <=> (x – 1)² + (y – 2)² + (z – 3)² ≤ 3 [(x – 1) + (y – 2) + (z – 3)]² ≤ 3[(x – 1)² + (y – 2)² + (z – 3)²] = 9 <=> – 3 ≤ (x – 1) + (y – 2) + (z – 3) ≤ 3 <=> 3 ≤ x + y + z ≤ 9 chắc đúng nhé bạn !!! áp dụng hằng đẳng thức là ra nhé !!! Bình luận
x² + y² + z² – 2x – 4y -6z ≤ -11
⇔ x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 ≤ 3
⇔ (x – 1)² + (y – 2)² + (z – 3)² ≤ 3
Mà [(x – 1) + (y – 2) + (z – 3)]² ≤ 3[(x – 1)² + (y – 2)² + (z – 3)²] = 9
⇒ – 3 ≤ (x – 1) + (y – 2) + (z – 3) ≤ 3
⇔ 3 ≤ x + y + z ≤ 9
x² + y² + z² – 2x – 4y -6z ≤ -11
<=> x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 ≤ 3
<=> (x – 1)² + (y – 2)² + (z – 3)² ≤ 3
[(x – 1) + (y – 2) + (z – 3)]² ≤ 3[(x – 1)² + (y – 2)² + (z – 3)²] = 9
<=> – 3 ≤ (x – 1) + (y – 2) + (z – 3) ≤ 3
<=> 3 ≤ x + y + z ≤ 9
chắc đúng nhé bạn !!! áp dụng hằng đẳng thức là ra nhé !!!