Chứng minh bất đẳng thức :

\(a^2+b^2+c^2\ge ab+bc+ac\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

\(\Leftrightarrow a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2\ge0\)\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

Đúng với mọi a, b, c

Ta có:

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc\)\(\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\left(1\right)\)\(\Leftrightarrow\left(a+b+c\right)^2\le a^2+b^2+c^2+2\left(ab+bc+ac\right)\)\(\Leftrightarrow\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(\Rightarrow a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{9}{3}=3\) Vậy GTNN của biểu thức là 3

Dấu "=" xảy ra khi và chỉ khi a=b=c

\(x^2+3x+7\)

\(=x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Vậy : GTNN của biểu thức là \(\dfrac{19}{4}\) khi \(x+\dfrac{3}{2}=0\Rightarrow x=\dfrac{-3}{2}\)

\(1,a^3+b^3-\left(a^2-2ab+b^2\right)\left(a-b\right)\)

\(\Leftrightarrow a^3+b^3-\left(a-b\right)^2\left(a-b\right)\)

\(\Leftrightarrow a^3+b^3-\left(a-b\right)^3\)

\(\Leftrightarrow a^3+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)

\(\Leftrightarrow a^3+b^3-a^3+3a^2b-3ab^2+b^3\)

\(\Leftrightarrow2b^3+3ab\left(a-b\right)\)

Tại a = -4 , b = 4

\(\Rightarrow2.4^3+3\left(-4\right)4\left(-4-4\right)=512\)

\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)=24\)

\(\Leftrightarrow6x-3-5x+15+18x-24-24=0\)

\(\Leftrightarrow19x-36=0\)

\(\Leftrightarrow19x=36\)

\(\Leftrightarrow x=\dfrac{36}{19}\)

\(a,A=a\left(a-6\right)+10=a^2-6a+9+1=\left(a-3\right)^2+1>0\)\(b,\left(x-3\right)\left(x-5\right)+4=x^2-8x+15+4=\left(x^2-8x+16\right)+3=\left(x-4\right)^2+3>0\)

\(c,a^2+a+1=\left(a^2+a+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\)

\(\left(-x+31\right)-39=69+11\)

\(\Leftrightarrow\left(-x+31\right)-39=80\)

\(\Leftrightarrow-x+31=80+39\)

\(\Leftrightarrow-x+31=119\)

\(\Leftrightarrow-x=119-31\)

\(\Leftrightarrow-x=88\Rightarrow x=-88\)

K là trung điểm của AC

\(\widehat{AKI}=\widehat{KCB}\) \(\Rightarrow\) IK // BC

\(\Rightarrow IK\) là đường trung bình của tam giác ABC:

\(\Rightarrow\) I là trung điểm của AB

\(\Rightarrow\) AI = 10cm

Vậy x = 10cm

**** cho mình đi các bạn

\(1,x^2+xy-8x-8y=x\left(x+y\right)-8\left(x+y\right)=\left(x-8\right)\left(x+y\right)\)\(3,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\)\(4,x^2-2xy+8x-16y=x\left(x-2y\right)+8\left(x-2y\right)=\left(x+8\right)\left(x-2y\right)\)\(5,x^2-9y^2+5x+15y=\left(x-3y\right)\left(x+3y\right)+5\left(x+3y\right)=\left(x+3y\right)\left(x-3y+5\right)\)