Question

1) Tìm x biết:

a)\(\sqrt{x+2}\) = \(\dfrac{5}{7}\)

b) \(\sqrt{x+2}\) - 8=1

c) 4- \(\sqrt{x-0,2}\) =0,5

2) Tìm giá trị nhỏ nhất của biểu thức sau:

a) A = \(\sqrt{x+24}\) +\(\dfrac{4}{7}\)

b)B = \(\sqrt{2x+\dfrac{4}{13}}\) - \(\dfrac{13}{191}\)

Answer 1

1)

a) \(\sqrt{x+2}=\dfrac{5}{7}\)

-> x+2 = \(\left(\dfrac{5}{7}\right)^{^2}\)=\(\dfrac{25}{49}\)

-> x = \(\dfrac{25}{49}-2=-\dfrac{73}{49}\)

b) \(\sqrt{x+2}-8=1\)

-> \(\sqrt{x+2}=1+8=9\)

-> \(x+2=9^2=81\)

-> x = 81 -2 = 79

c) 4 - \(\sqrt{x-0,2}=0,5\)

-> \(\sqrt{x-0,2}=4-0,5=3,5\)

-> x - 0,2 = (3,5)² = 12,25

-> x = 12,25 +0,2 = 12,45

2) a)

Với mọi x thì: \(\sqrt{x+24}\ge0\)

=> \(\sqrt{x+24}+\dfrac{4}{7}\ge\dfrac{4}{7}\)

Dấu "=" xảy ra khi : x + 24 = 0 <=> x = -24

Vậy Min_A = \(\dfrac{4}{7}\) khi x = -24