\(1,\\ a,=A^2+2AB+B^2\\ b,=\left(A-B\right)\left(A+B\right)\\ 2,\\ a,=\left(17-7\right)^2=10^2=100\\ b,=\left(2021-2020\right)\left(2021+2020\right)=4041\\ 3,\\ a,=3\left(x-2\right)\\ b,=\left(x-2\right)^2\\ c,=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)=\left(x+y+5\right)\left(x-y\right)\)
\(d,\) Đặt \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\x+z=c\end{matrix}\right.\Rightarrow x+y+z=\dfrac{a+b+c}{2}\)
Thế vào đề:
\(BT=8\cdot\dfrac{\left(a+b+c\right)^3}{8}-a^3-b^3-c^3\\ =\left(a+b+c\right)^3-a^3-b^3-c^3\\ =a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3\\ =3\left(a+b\right)\left(b+c\right)\left(c+a\right)\\ =3\left(2x+y+z\right)\left(x+2y+z\right)\left(x+y+2z\right)\)
\(4,\\ a,=6a^2+3ab\\ b,=2a^3-6a+3\\ 5,\\ a,=\left(3x-1\right)^2+y^2+4\ge4\\ Dấu"="\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow2x^3-3x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\)
Thay \(x=-1\Leftrightarrow-2-3-1+a=0\Leftrightarrow a=6\)
