\(a,21bc^2-6c-3c^3+42b\)

\(=21bc^2-3c^3+42b-6c\)

\(=3c^2\left(7b-3c\right)+6\left(7b-3c\right)=3\left(c^2+2\right)\left(7b-3c\right)\)\(c,ax-bx-cx+ay-by-cy\)

\(=x\left(a-b-c\right)+y\left(a-b-c\right)=\left(x+y\right)\left(a-b-c\right)\)Còn phần b hình như sai đề

\(a,x^2-7x-30=x^2-10x+3x-30\)

\(=x\left(x-10\right)+3\left(x-10\right)\)

\(=\left(x-10\right)\left(x+3\right)\)

\(b,x^2-x-6=x^2-3x+2x-6\)

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

\(c,x^2+x-30=x^2-5x+6x-30\)

\(=x\left(x-5\right)+6\left(x-5\right)=\left(x+6\right)\left(x-5\right)\)

\(a,4x^2+5x+3\)

\(=\left(4x^2+5x+\dfrac{25}{16}\right)+\dfrac{23}{16}\)

\(=\left(2x+\dfrac{5}{4}\right)^2+\dfrac{23}{16}\)

Với mọi giá trị của x ta có:

\(\left(2x+\dfrac{5}{4}\right)^2\ge0\Rightarrow\left(2x+\dfrac{5}{4}\right)^2+\dfrac{23}{16}\ge0\)

=>đpcm

b, \(7x^2-x+8=7\left(x^2-\dfrac{1}{7}x+\dfrac{1}{196}\right)+\dfrac{223}{28}\)

\(=\left(x-\dfrac{1}{14}\right)^2+\dfrac{223}{28}\)

Với mọi giá trị của x ta có:

\(\left(x-\dfrac{1}{14}\right)^2\ge0\Rightarrow\left(x-\dfrac{1}{14}\right)^2+\dfrac{223}{28}\ge0\)

=> đpcm

\(c,25x^2+8x+2017=25\left(x^2+\dfrac{8}{25}x+\dfrac{16}{625}\right)+\dfrac{50409}{25}\)\(=\left(x+\dfrac{4}{25}\right)^2+\dfrac{50409}{25}\)

Với mọi giá trị của x ta có:

\(\left(x+\dfrac{4}{25}\right)^2\ge0\Rightarrow\left(x+\dfrac{4}{25}\right)^2+\dfrac{50409}{25}\ge0\)=>đpcm

\(b,y^2\left(x^2+y\right)-x^2z-yz\)

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

\(c,3x\left(x+1\right)^2-5x^2\left(x+1\right)+7x+7\)

\(=3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)

\(=\left(x+1\right)\left[3x\left(x+1\right)-5x^2+7\right]\)

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

\(d,x^3-27+x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+4x+9\right)\)

\(4x^2+5x+3\)

\(=\left(4x^2+5x+\dfrac{25}{16}\right)+\dfrac{23}{16}\)

\(=\left(2x+\dfrac{5}{4}\right)^2+\dfrac{23}{16}\)

Với mọi giá trị của x ta có:

\(\left(2x+\dfrac{5}{4}\right)^2\ge0\Rightarrow\left(2x+\dfrac{5}{4}\right)^2+\dfrac{23}{16}\ge0\)

=> đpcm

\(x-y=5\Rightarrow\left(x-y\right)^2=25\)

\(\Leftrightarrow x^2-2xy+y^2=25\)

\(\Leftrightarrow-2xy=25-\left(x^2+y^2\right)\)

\(\Leftrightarrow-2xy=25-15=10\)

\(\Rightarrow xy=-5\)

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

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

\(=5.\left[15+\left(-5\right)\right]=50\)

Ta có:

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

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

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

\(\Leftrightarrow a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2=0\)\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)

\(,A=x^2-12x+37=\left(x^2-12x+36\right)+1\)

\(=\left(x-6\right)^2+1\)

với mọi giá trị của x , ta có:

\(\left(x-6\right)^2\ge0\Rightarrow\left(x-6\right)^2+1\ge1\)

Vậy Min A = 1

Để A = 1 thì \(x-6=0\Rightarrow x=6\)

\(B=-x^2+14x-53\)

\(=-\left(x^2-14x+49\right)-4\)

\(=-\left(x-7\right)^2-4\le-4\)

Vậy Max B = -4

Để B = -4 thì \(x-7=0\Rightarrow x=7\)

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

\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)

\(=2x\left(4x+2\right)=4x\left(2x+1\right)\)