\(a.x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)

\(b.x^2+8x+12=x^2+2x+6x+12=x\left(x+2\right)+6\left(x+2\right)=\left(x+6\right)\left(x+2\right)\)

\(c.x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\)

\(d.x^2-x+6\) (Vì pt này vô nghiệm nên ko tách đc thành nhân tử)

\(A=2020+2021+2022+2023+...+2040\)

\(=\left(2020+2040\right).10=40600\)

\(B=587+125+460+875+540+413\)

\(=\left(587+413\right)+\left(125+875\right)+\left(460+540\right)\)

\(=1000+1000+1000=3000\)

\(a.9-25x^2=\left(3+5x\right)\left(3-5x\right)\)

\(b.4-4x+x^2=\left(2-x\right)^2\)

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

\(d.16-\left(x+1\right)^2=\left(x+5\right)\left(3-x\right)\)

\(e.25-9\left(x-2\right)^2=\left[5+3\left(x-2\right)\right]\left[5-3\left(x-2\right)\right]=\left(3x-1\right)\left(11-3x\right)\)

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

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

\(x^2-2x-1>0\)

\(\Leftrightarrow-\left(x^2+2x+1\right)>0\)

\(\Leftrightarrow-\left(x+1\right)^2>0\left(vô.lí\right)\)

Vậy BPT vô nghiệm

ha = hm²

m không đổi ra đc hm² đâu nhé

là số tự nhiên thôi nên chỉ có 4TH

4 trường hợp e nhé

\(TH_1:x=1;y=2\)

\(TH_2:x=2;y=1\)

\(TH_3:x=0;y=3\)

\(TH_4:x=3;y=0\)

Bình phương 2 vế em nhé

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{m}{2}\\x_1x_2=\dfrac{m-2}{2}\end{matrix}\right.\)

\(\Delta=m^2-8m+16=\left(m-4\right)^2\ge0\forall m\)

\(x_1-x_2=x_1x_2\)

\(\Leftrightarrow\left(x_1-x_2\right)^2=\left(\dfrac{m-2}{2}\right)^2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=\dfrac{m^2-4m+4}{4}\)

\(\Leftrightarrow\dfrac{m^2}{4}-\dfrac{4m-8}{2}-\dfrac{m^2-4m+4}{4}=0\)

\(\Leftrightarrow\dfrac{m^2-8m+16-m^2+4m-4}{4}=0\)

\(\Leftrightarrow\dfrac{-4m+8}{4}=0\)

\(\Leftrightarrow-m+2=0\)

\(\Leftrightarrow m=2\)