\(\left(4n+3\right)^2-25\)

\(=\left(4n+3\right)^2-5^2\)

\(=\left(4n+3-5\right)\left(4n+3+5\right)\)

\(=\left(4n-2\right)\left(4n+8\right)\)

xl chia hết cho 8

\(A=x^2-4x+4+y^2-8y+16-14=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

giúp tui vs

huhu

\(A=-2x^2+x+5\)

\(=-2\left(x^2-x-5\right)\)

\(=-2\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{19}{4}\right)\)

\(=-2\left(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\right)\)

\(=-\frac{19}{2}-2\left(x-\frac{1}{2}\right)^2\le\frac{-19}{2}\)

MIN \(A=\frac{-19}{2}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

\(\left(x-1\right)\left(x+2\right)-\left(x+1\right)x\)

\(=x^2+x-2-x^2-x\)

\(=\left(x^2-x^2\right)+\left(x-x\right)-2\)

\(=-2\)

Giá trị rút gọn là:

 (x-1)*(x+2)-x*(x+1)

  Đáp số: (x-1)*(x+2)-x*(x+1)

• Theo bất đẳng thức tam giác , ta có : \(AO+OB>AB\)

\(OB+OC>BC\)

\(OC+OD>CD\)

\(OD+OA>AD\)

\(\Rightarrow2\left(OA+OB+OC+OD\right)>AB+BC+CD+DA\Leftrightarrow AC+BD>\frac{AB+BC+CD+DA}{2}\)

• Tương tự, ta có : \(AC< AB+BC\) ; \(AC< AD+CD\)

\(BD< AB+AD\) ; \(BD< BC+CD\)

\(\Rightarrow2\left(AC+BD\right)< 2\left(AB+BC+CD+AD\right)\Leftrightarrow AC+BD< AB+BC+CD+AD\)

Vậy ta có : \(\frac{AB+BC+CD+AD}{2}< AC+BD< AB+BC+CD+AD\)