Question

Bài 1: Phân tích đa thức thành nhân tử:

a) x(x+y)-6x-6y

b)x^3-x-y62-y

c)2x^3+16

d)x^4+2x^3+x^2

e)5x^2-10xy+5y^2-20z^2

f)16-x^2+2xy-y^2

G)3x^2y-6x^2y+3y

Bài 2: tìm x:

a)x^2-10x=-25

b) x^2(x-2)+x-2=0

c) (x-2)^2=(3x-5)^2

Answer 1

giải hết nha đừng có ghi kết quả

Answer 2

Bài 1 :

a,\(x\left(x+y\right)-6x-6y=x\left(x+y\right)-6\left(x+y\right)=\left(x+y\right)\left(x-6\right)\)

b, \(x^3-x-y^2-y??????????\)

c, \(2x^3+16=2\left(x^3+8\right)=2\left(x+3\right)\left(x^2-3x+9\right)\)

d, \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

e, \(5x^2-10xy+5y^2-20z^2=5[\left(x-y\right)^2-\left(2z\right)^2]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

f, \(16-x^2+2xy-y^2=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

g, Tương tự e

Answer 3

Bài 2 :

a,\(x^2-10x=-25< =>\left(x-5\right)^2=0< =>x=5\)

b,\(x^2\left(x-2\right)+x-2=0\)

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

\(x^2+1>0< =>x-2=0< =>x=2\)

c, \(\left(x-2\right)^2=\left(3x-5\right)^2\)

\(< =>\left[{}\begin{matrix}x-2=3x-5\\x-2=5x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=0,25\end{matrix}\right.\)