1, Tìm x biết:
a, x\(^2\) - 2x +1 = 0
b, ( 5x + 1)\(^2\) - (5x - 3) ( 5x + 3) = 30
c, ( x - 1) ( x\(^2\) + x + 1) - x ( x +2 ) ( x - 2) = 5
d, ( x - 2)\(^3\) - ( x - 3) ( x\(^2\) + 3x + 9 ) + 6 ( x + 1)\(^2\) = 15
2, Chứng minh các đa thức sau luôn luôn dương với mọi x, y :
a, x\(^2\) + 2x + 2
b, 4x\(^2\) - 12x + 11
c, x\(^2\) - x + 1
d, x\(^2\) - 2x + y\(^2\) + 4y + 6
3, Tìm x, y biết :
a, x\(^2\) + y\(^2\) - 2x + 4y + 5 = 0
b, x\(^2\) + 4y\(^2\) + 6x - 12y + 18 = 0
c, 5x\(^2\) + 9y\(^2\) - 12xy - 6x + 9 = 0
d, 2x\(^2\) + 2y\(^2\) + 2xy - 10x - 8y + 41 = 0
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
2, Chứng minh các đa thức sau luôn luôn dương với mọi x, y :
a, x2 + 2x + 2
= x2 + 2x +1 +1
= (x+1)2 +1
Vì (x+1)2 \(\ge\) 0 , \(\forall\) x
nên (x+1)2 +1 \(\ge\) 1 > 0 , \(\forall\) x
b, 4x2 - 12x + 11
= (2x)2 - 2.2x.3 + 9 +2
= (2x - 3)2 +2
Vì (2x - 3)2 \(\ge\)0, \(\forall\)x
nên (2x - 3)2 +2 \(\ge\)2 >0 , \(\forall\) x
c, x2 - x + 1
= x2 - 2.x . \(\dfrac{1}{2}\) +\(\dfrac{1}{4}\)+ \(\dfrac{3}{4}\)
= (x - \(\dfrac{1}{2}\))2 + \(\dfrac{3}{4}\)
Vì (x - \(\dfrac{1}{2}\))2 \(\ge\) 0, \(\forall\) x
nên (x - \(\dfrac{1}{2}\))2 + \(\dfrac{3}{4}\) \(\ge\) \(\dfrac{3}{4}\) \(\ge\)0, \(\forall\) x
d, x2 - 2x + y2 + 4y + 6
= (x2 - 2x + 1)+ (y2 + 4y + 4) +1
= (x-1)2 + (y+2)2 +1
Vì (x-1)2 \(\ge\) 0, \(\forall\) x
(y+2)2 \(\ge\)0, \(\forall\) y
nên (x-1)2 + (y+2)2 +1 \(\ge\)1 >0, \(\forall\) x, y
