ta có :

\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)-1\)

\(=\left(x-y\right)^2+\left(x+1\right)^2-1\ge-1\). Vậy GTNN của A =-1 khi \(x=y=-1\).

\(B=\frac{1}{2}\left[\left(a^2-2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\right]\)

\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\right]\ge0\) Vậy GTNN của B =0 khi \(a=b=-c\)

ta có :

\(7x^3-14x^2+7x=7x\left(x^2-2x+1\right)=7x\left(x-1\right)^2\)

ta có:

Hình tự vẽ nha.

Lời giải:

+ Xét\(\Delta AHB\)và\(\Delta AKC\)có:

\(\widehat{AHB}=\widehat{AKC}=90^0\)

\(AB=AC\)(Do\(\Delta ABC\)cân tại A)

\(\widehat{HAB}=\widehat{KAC}\)

Do đó:\(\Delta AHB=\Delta AKC\)(g-c-g)

\(\Rightarrow AH=AK\)

\(\Rightarrow\Delta AHK\)cân tại A

\(\Rightarrow\widehat{AKH}=\frac{180^0-\widehat{A}}{2}\)

Mà\(\widehat{ABC}=\frac{180^0-\widehat{A}}{2}\)(Do\(\Delta ABC\)cân tại A)

\(\Rightarrow\widehat{AKH}=\widehat{ABC}\)

\(\Rightarrow HK//BC\)

+Xét tứ giác BCKH có\(HK//BC\)

=> BCHK là hình thang

Mà\(\widehat{B}=\widehat{C}\)(Do\(\Delta ABC\)cân tại A)

=> BCHK là hình thang cân (đpcm)

Vậy BCHK là hình thang cân

\(1,a,A=x^2-6x+25\)

\(=x^2-2.x.3+9-9+25\)

\(=\left(x-3\right)^2+16\)

Ta có :

\(\left(x-3\right)^2\ge0\)Với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

Hay \(A\ge16\)

\(\Rightarrow A_{min}=16\)

\(\Leftrightarrow x=3\)

\(b,B=4x^2+4x-2\)

\(B=4x^2+4x+1-3\)

\(B=\left(4x^2+4x+1\right)-3\)

\(B=\left(2x+1\right)^2-3\)

Ta có : 

\(\left(2x+1\right)^2\ge0\)với mọi x

\(\Rightarrow\left(2x+1\right)^2-3\ge-3\)

\(\Leftrightarrow B\ge-3\)

\(\Rightarrow B_{min}=-3\)

\(\Leftrightarrow x=-\frac{1}{2}\)

a mũ 3 + 8 + 2 a mũ 2

a) \(18x^2y-12x^3\)

\(=6x^2.3y-6x^2.2x\)

\(=6x^2\left(3y-2x\right)\)

b) \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(5x+3\right)\left(x-y\right)\)

Phần 3) đề thiếu, bạn bổ sung nhé.

\(4x^2-12x=4x.\left(x-3\right)\)

\(8x.\left(x-1\right)+10.\left(1-x\right)=8x.\left(x-1\right)-10.\left(x-1\right)=\left(8x-10\right).\left(x-1\right)=2.\left(4x-5\right).\left(x-1\right)\)

\(\left(x+2\right).\left(x+4\right).\left(x+6\right).\left(x+8\right)+16\)

\(=[\left(x+2\right).\left(x+8\right)].[\left(x+4\right).\left(x+6\right)]+16\)

\(=\left(x^2+2x+8x+16\right).\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right).\left(x^2+10x+24\right)+16\)

Ta đặt: \(n=x^2+10x+16\)

\(=n.\left(n+8\right)+16\)

\(\Rightarrow n^2+8n+16\)

\(\Rightarrow n^2+2n.4+4^2\)

\(=\left(n+4\right)^2\)

Ta thay \(n=x^2+10x+16\)

\(=\left(x^2+10x+16+4\right)^2\)

\(=\left(x^2+10+20\right)^2\)

ta có :

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

\(=1-2ab\ge1-\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)

dấu bằng xảy ra khi \(a=b=\frac{1}{2}\)

\(P=a^3+b^3+ab\)

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

\(=1-3ab+ab=1-2ab\)

\(\ge1-\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)