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

\(a^4+a^3+a^2+a\)

\(=a^3\left(a+1\right)+a\left(a+1\right)\)

\(=\left(a+1\right)\left(a^3+a\right)\)

nha !!!

Trả lời:

\(a^4+a^3+a^2+a\)

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

\(=a^3\left(a+1\right)+a\left(a+1\right)\)

\(=a\left(a+1\right)\left(a^2+1\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(a^4-a^3-a^2+a\)

\(=a^3\left(a-1\right)-a\left(a-1\right)\)

\(=\left(a-1\right)\left(a^3-a\right)\)

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

Để đơn giản hơn cũng như là dễ nhìn hơn thì ta :

Đặt : \(x^2+2x=a\)

Do đó ta có đa thức :

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

\(=a^2+a+3a+3\)

\(=a\left(a+1\right)+3\left(a+1\right)\)

\(=\left(a+1\right)\left(a+3\right)\)

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

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

Hoặc bạn có thể đặt \(x^2+2x+2=t\)

Thì \(P=\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)

\(P=\left(t-2\right)\left(t+2\right)+3\)

\(P=t^2-4+3\)

\(P=t^2-1\)

\(P=\left(t-1\right)\left(t+1\right)\)

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

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

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

\(=x^4+4x^3+4x^2+4x^3+16x^2+16x\)

\(=x^4+8x^3+20x^2+16x\)

\(=\left(x^4+8x^3+20x^2+16x\right)+3\)

\(=x^4+8x^3+20x^2+16x+3\)

a) (x + 1)(x + 2)(x + 3)(x + 4) - 24

= [(x + 1)(x + 4)].[(x + 2)(x + 3)] - 24

= (x² + 5x + 4)(x² + 5x + 6) - 24 

= (x² + 5x + 5 - 1)(x² + 5x + 5 + 1) - 24

= (x² + 5x + 5)² - 1 - 24 = (x² + 5x + 5)² - 25 

= (x² + 5x)(x² + 5x + 10) 

 = x(x + 5)(x² + 5x + 10)

=a^4(a+1)+a^2(a+1)+(a+1)

=(a+1)(a^4+a^2+1)

a⁵+a⁴+a³+a²+a+1

=a⁴(a+1)+a²(a+1)+(a+1)

=(a+1)(a⁴+a²+1)

 1)a)x(x+4)(x-4)-(x2+1)(x2-1) 
=x(x2-16)-(x^4-1)=-x^4+x^3-16x+1 
b)4(x-1)(x+1)-(x+2)(x+5)-3(x-1)(x+2) 
=4(x2-1)-(x2+7x+10)-3(x2+x-2) 
=4x2-4-x2-7x-10-3x2-3x+6 
=-10x-8 
2)a)x2-2x-4y2-4y 
=(x2-4y2)-(2x+4y) 
=(x-2y)(x+2y)-2(x+2y) 
=(x+2y)(x-2y-2) 
b)x2-x-2=x2-2x+x-2 
=x(x-2)+(x-2)=(x-2)(x+1) 
c)a5+a+1=a5+a4+a3-a4-a3-a2+a2+a+1 
=a3(a2+a+1)-a2(a2+a+1)+(a2+a+1) 
=(a2+a+1)(a3-a2+1) 
3)a)x3-4x2-8x+8 
=(x3+8)-(4x2+8x) 
=(x+2)(x2-2x+4)-4x(x+2) 
=(x+2)(x2-6x+4) 
b)4x4+1 
=4x4+4x2+1-4x2 
=(2x2+1)^2 -(2x)^2 
=(2x2+1+2x)(2x2+1-2x) 
c)x2(x2+4)-x2+4 
=x4-x2+4x2+4 
=x4+4x2+4-x2 
=(x2+2)^2-x^2 
=(x2-x+2)(x2+x-2) 
=(x2-x+2)(x2+2x-x-2) 
=(x2-x+2)(x+2)(x-1) 
4)a)Bạn đặt phép chia (2x2+ax+1)cho(x-3)thì ta được số dư là :3a+19(bài dài quá nên mình không muốn đánh bạn tự chia nhé).mà để phép chia dư 4 thì 3a+19=4 =>a=-5 
vậy a=-5 
b)ta đặt:g(x)= ax5+5x4-x=(x-1).f(x) 
g(1)=a+4=0.f(x)=0 =>a=-4 
Bạn làm như anh Hồ Công Tử là tốt lắm rồi 
5)a)(5-2x)(2x+7)=4x2-25 
<=>(5-2x)(2x+7)+(25-4x2)=0 
<=>(5-2x)(2x+7)+(5-2x)(5+2x)=0 
<=>(5-2x)(2x+7+5+2x)=0 
<=>(5-2x)(4x+12)=0 
<=>5-2x=0 hoặc 4x+12=0 
<=>x=2,5 hoặc x=-3 
b)x3+27+(x+3)(x-9)=0(bạn hãy chú ý đến 3^3=27và a^3+b^3=(a+b)(a2-ab+b2)) 
<=>(x+3)(x2-3x+9)+(x+3)(x-9)=0 
<=>(x+3)(x2-2x)=0 
<=>x(x+3)(x-2)=0 
<=>x=0 hoặc x+3=0 hoặc x-2=0 
<=>x=0 hoặc x=-3 hoặc x=2 
c)x2+3x-18=0 
<=>x2+6x-3x-18=0 
<=>x(x+6)-3(x+6)=0 
<=>(x+6)(x-3)=0 
<=>x+6=0 hoặc x-3=0 
<=>x=-6 hoặc x=3 

k mình nha mấy bạn

1: \(\dfrac{A}{B}=\dfrac{2x^4+4x^3-x^3-2x^2-2x^2-4x+x+2}{x+2}\)

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