Câu 1: C

Câu 2: A

Câu 3: B

Câu 4:B

Câu 5: A

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

A = (3x-2)^2-(x+3)^2

= 9x^2 - 12x + 4 - x^2 - 6x - 9

= 8x^2 - 18x - 5

B = (5x+3)^2+(x-2)^2

= 25x^2 + 30x + 9 + x^2 - 4x + 4

= 26x^2 +26x +13

C = (2x+y-3)^2-(x+2y+3)^2

= (2x + y)^2 - 6(2x + y) + 9 - (x + 2y)^2 - 6(x + 2y) - 9

= 4x^2 + 4xy + y^2 - 12x - 6y - x^2 - 4xy - 4y^2 - 6x - 12y 

= 3x^2 - 3y^2 -18x - 18y

D = (x+2y+3z)^2 -(x-2y-3z)^2

= (x + 2y)^2 + 6z(x + 2y) + 9z^2 - (x - 2y)^2 + 6z(x - 2y) - 9z^2

= x^2 + 4xy + y^2 + 6xz + 12yz - x^2 + 4xy - y^2 + 6xz - 12yz 

= 8xy + 12xz 

A= (3x-2)^2-(x+3)^2

= (9x^2 - 12x + 4) - (x^2 + 6x +9)

= 8x^2 - 6x + 13

Chết rùi! Quên đổi dấu => làm sai/

Đặt \(m=3k+r\)với \(0\le r\le2\)        \(n=3t+s\)với \(0\le s\le2\)

\(\Rightarrow x^m+x^n+1=x^{3k+r}+x^{3t+s}+1=x^{3k}+x^r-x^r+x^{3t}x^s-x^s+x^r+x^s+1\)

\(=x^r\left(x^{3k}-1\right)+x^s\left(x^{3t}-1\right)+x^r+x^s+1\)

Ta thấy : \(\left(x^{3k}-1\right)⋮\left(x^2+x+1\right)\)và \(\left(x^{3t}-1\right)⋮\left(x^2+x+1\right)\)

Vậy : \(\left(x^m+x^n+1\right)⋮\left(x^2+x+1\right)\)

\(\Leftrightarrow\left(x^r+x^s+1\right)⋮\left(x^2+x+1\right)\)với \(0\le r;s\le2\)

\(\Leftrightarrow\hept{\begin{cases}r=2\\r=1\end{cases}}\)và\(\hept{\begin{cases}s=1\\s=2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}m=3k+2\\m=3k+1\end{cases}}\)và\(\hept{\begin{cases}n=3t+1\\n=3t+2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}mn-2=\left(3k+2\right)\left(3t+1\right)-2=9kt+3k+6t=3\left(3kt+k+2t\right)\\mn-2=\left(3k+1\right)\left(3t+2\right)-2=9kt+6k+3t=3\left(3kt+2k+t\right)\end{cases}}\)

\(\Leftrightarrow\left(mn-2\right)⋮3\)Điều phải chứng minh 

Áp dụng : \(m=7;n=2\Rightarrow mn-2=12:3\)

\(\Rightarrow\left(x^7+x^2+1\right)⋮\left(x^2+x+1\right)\)

\(\Rightarrow\left(x^7+x^2+1\right):\left(x^2+x+1\right)=x^5+x^4+x^2+x+1\)

\(7,D\\ 8,D\\ 9,B\\ 10,B\\ 11,B\\ 12,C\\ 13,B\\ 14,A\)

tao vả bay lồn mày ấy hiểu chưa toán lớp 1 mà như cái phương chình đại học vậy hả thằng lồn ngu này

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

\(8x^3+36x^2y+54xy^2+27y^3\\ =\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\\ =\left(2x+3y\right)^3\\ =\left(2x+3y\right)\left(2x+3y\right)\left(2x+3y\right)\)

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

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

\(\left(x-1\right)^2-\left(x+1\right)^2\\ =\left(x-1-x-1\right)\left(x-1+x+1\right)\\ =-2.2x=-4x\)

a: =(2x)^3+3*(2x)^2*3y+3*2x*(3y)^2+(3y)^3

=(2x+3y)^3

b: (x-y)^3-(x+y)^3

=(x-y-x-y)[(x-y)^2+(x-y)(x+y)+(x+y)^2]

=-2y*[x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2]

=-2y(3x^2+y^2)

c: (x+1)^3+(x-1)^3

=(x+1+x-1)[(x+1)^2-(x+1)(x-1)+(x-1)^2]

=2x*[x^2+2x+1-x^2+1+x^2-2x+1]

=2x(x^2+3)

d: =(x-1-x-1)(x-1+x+1)

=2x*(-2)=-4x

\(8x^3+36x^2y+54xy^2+27y^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)

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

\(------\)

\(\left(x-y\right)^3-\left(x+y\right)^3\)

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

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

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

\(------\)

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

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

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

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

\(------\)

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

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

\(=-2.2x=-4x\)