a ) Vì | x + 1 | ≥ 0 ∀ x ∈ N
Để A = | x + 1 | + 1,7 min <=> x + 1 = 0 => x = - 1
Vậy min A = 1,7 <=> x = - 1
b ) Vì B = | x - 2/3 | ≥ 0 ∀ x ∈ N
Để | x -2/3 | + 3/7 min <=> x - 2/3 = 0 => x = 2/3
Vậy min B = 3/7 <=> x = 2/3
tim GTNN cua bieu thuc :
a) A= I x+1I + 1,7
b) B= I x-2/3I +3/7
c) C= 5+ -8/ 4x I5x+7I 24
tim gia tri lon nhat cua bieu thuc :
a) C= 5+ 15/ 4 I 3x+7 I +3
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
a) A= I x+1I + 1,7
b) B= I x-2/3I +3/7
c) C= 5+ -8/ 4x I5x+7I 24
tim gia tri lon nhat cua bieu thuc :
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
c) C= 5+ -8/ 4x I5x+7I 24
tim gia tri lon nhat cua bieu thuc :
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
c) C= 5+ -8/ 4x I5x+7I 24
Tìm GTNN của:
a) A= I x - 7 I + 3 - x
b) B= I x+ 7I + I x + 3I + I x + 1I
c) C= I 2 - xI + I 5 - xI
Tim GTNN cua bieu thuc A= 4x^2 + 12x + 8
a,Tim GTNN cua bieu thuc \(C=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\)
b,Tim GTLN cua bieu thuc \(D=\frac{4}{\left(2x-3\right)^2+5}\)
bài 1:Tìm x:
a)2.I2x-3I=\(\frac{1}{2}\)
b)7,5-3.I5-2xI=-4,5
c)I7x-1I-I5x+6I=0
d)I\(\frac{5}{4}\).x-\(\frac{7}{2}\)I-I\(\frac{5}{8}\).x+\(\frac{3}{5}\)I=0
tim gtnn cua bieu thuc:
A=|x|+|8-x|
