1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

Tôi làm tạm theo cách này nhé.

\(x^4-2x^2-114x-1295\)

\(=\frac{d}{dx}\left(x^4-2x^2-114x-1295\right)\)

\(=4x^3-4x-114-0\)

\(=4x^3-4x-114\)

Bạn Phương Lê Nhật ơi!!!!

Đây là Toán 8 bạn ạ

Bạn giải mk ko hiểu j cả

Giải cụ thể đc ko bạn ạ

a: =64x^4+16x^2y^2+y^4-16x^2y^2

=(8x^2+y^2)^2-(4xy)^2

=(8x^2+y^2-4xy)(8x^2+y^2+4xy)

b: =x^8+2x^4+1-x^4

=(x^4+1)^2-x^4

=(x^4-x^2+1)(x^4+x^2+1)

=(x^4-x^2+1)(x^4+2x^2+1-x^2)

=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)

c: =(x+1)(x^2-x+1)+2x(x+1)

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

=(x+1)(x^2+x+1)

d: =(x^2-1)(x^2+1)-2x(x^2-1)

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

=(x-1)^2*(x-1)(x+1)

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

\(x^4-y^2\left(2x-y\right)^2\)

\(=x^4-\left(2xy-y^2\right)^2\)

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

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)