a, Ta có OB = \(\dfrac{1}{2}\)OC = \(5.\dfrac{1}{2}=2,5\left(cm\right)\)

=>  B nằm giữa O và C

b, \(BC=OC-OB=5-2,5=2,5\left(cm\right)\)

=> OB = BC= 2,5 (cm)

c, Ta có OB = BC =2,5 (cm)

=> B là trung điểm OC

exam

\(x^2+2x+3x+6-\left(x^2-2x+5x-10\right)=0\\ x^2+5x+6-x^2-3x+10=0\\ 2x+16=0\\ 2x=-16\\ x=\left(-16\right):2\\ x=-8\)

A

\(a,B=12-3x^2+4x^2+4x+1-2x^2-6=-x^2+4x+7\)

b, \(-\left(x^2-4x-7\right)=-\left(x^2-4x+4\right)+11=-\left(x-2\right)^2+11\)

\(\left(x-2\right)^2\ge0\\ -\left(x-2\right)^2\le0\\ -\left(x-2\right)^2+11\le11\)

Dấu " =" xảy ra khi

\(x-2=0\\ x=2\)

Vậy GTLN B là : 11 khi x =2

\(x^2-3x-2x+6-\left(x^2+x-5x-5\right)=0\\ x^2-5x+6-x^2+4x+5=0\\ -x+11=0\\ -x=-11\\ x=11\)

\(a,x^2-\dfrac{1}{9}=0\\ x^2=\dfrac{1}{9}\\\)

x= \(\dfrac{1}{3};x=-\dfrac{1}{3}\)

\(b,x^2+2x+1=0\\ \left(x+1\right)^2=0\\ x+1=0\\ x=-1\)

\(=x^2-x+2020x-2020-x-2019=x^2-2018x-4039\)

\(a.2^2-a.2+2=2\\ 4a-2a+2=2\\ 2a+2=2\\ 2a=0\\ a=0\)

dialogue