Bài 7
\(x^4 + y^4 + ( x+y)^4
\)
=\(x^4 + y^4 + ( x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4)\)
=\(2x^4 + 2y^4+ 4x^3y + 6x^2y^2 + 4xy^3\)
=2\((x^4 + y^4 +2x^3y + 3x^2y^2 + 2xy^3)\)
=2(\(x^2 + xy + y^2)^2\) ( ĐPCM)
Bài 8
a) \(9x^2 - 6x + 2\)
\(= (9x^2 - 6x + 1 ) + 1\)
\(= (3x-1)^2 + 1 \)
Vì \((3x-1)^2 \) ≥ 0
→ \((3x-1)^2 \) + 1 ≥ 1 (ĐPCM)
b)\(x^2 + x + 1\)
\(=(x^2 + x + \dfrac{1}{4}) + \dfrac{3}{4}
\)
=\((x + \dfrac{1}{2} )^2 + \dfrac{3}{4}\) ≥ \(\dfrac{3}{4}
\) ( ĐPCM )
c)\(2x^2 + 2x + 1\)
\(= x^2 + (x^2 + 2x + 1)\)
\(= x^2 + (x+1)^2\) > 0 ( ĐPCM )
Bài 9
a) A = \(x^2 - 3x + 5\)
⇔A = \(x^2 - 3x +\dfrac{9}{4}\) +\(\dfrac{11}{4}\)
⇔A = \((x - \dfrac{3}{2})^2\) + \(\dfrac{11}{4}\) ≥ \(\dfrac{11}{4}\)
Vậy A min = \(\dfrac{11}{4}\) khi x = \(\dfrac{3}{2}\)
b) B = \((2x-1)^2 + (x+2)^2\)
⇔ B = \(4x^2 - 4x + 1 + x^2 + 4x + 4\)
⇔ B = \(5x^2 + 5\) ≥ 5
Vậy B min = 5 khi x = 0
Bài 7
x4+y4+(x+y)4x4+y4+(x+y)4
=x4+y4+(x4+4x3y+6x2y2+4xy3+y4)x4+y4+(x4+4x3y+6x2y2+4xy3+y4)
=2x4+2y4+4x3y+6x2y2+4xy32x4+2y4+4x3y+6x2y2+4xy3
=2(x4+y4+2x3y+3x2y2+2xy3)(x4+y4+2x3y+3x2y2+2xy3)
=2(x2+xy+y2)2x2+xy+y2)2 ( ĐPCM)
Bài 8
a) 9x2−6x+29x2−6x+2
=(9x2−6x+1)+1=(9x2−6x+1)+1
=(3x−1)2+1=(3x−1)2+1
Vì (3x−1)2(3x−1)2 ≥ 0
→ (3x−1)2(3x−1)2 + 1 ≥ 1 (ĐPCM)
b)x2+x+1x2+x+1
(x+12)2+34(x+12)2+34 ≥
