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

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

a) \(x^4-2x^2=0\)

⇔ \(x^2\left(x^2-2\right)=0\)

⇔ \(x^2\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)

⇔ \(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)

b) \(x^2+3x+2=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

1) \(5x-5y+x\left(x-y\right)\)

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

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

2) \(x^2+4x+3\)

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

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

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

3) \(x^2-2xy+y^2-z^2\)

\(=\left(x-y\right)^2-z^2\)

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

4) \(x\left(x-5\right)-3x+15\)

\(=x\left(x-5\right)-3\left(x-5\right)\)

\(=\left(x-5\right)\left(x-3\right)\)

5) \(y^2-x^2+2x-1\)

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

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

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

\(1.\left(x-y\right)\left(x+5\right)\)

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

\(3.\left(x-y-z\right)\left(x-y+z\right)\)

\(4.\left(x-3\right)\left(x-5\right)\)

\(5.\left(y-x+1\right)\left(y+x+1\right)\)

\(7.\left(x+1\right)\left(x-2\right)^2\)

\(8.\left(x-5\right)\left(x+3\right)\)

\(10.\left(y+1\right)\left(2x+z\right)\)

1)

5x - 5y + x ( x - y ) = (x-y)(5+x)

2)

x²+4x+3=x²+x+3x+3=(x+1)(x+3)

3)x²-2xy+y²-z²=\(\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

4)\(x\left(x-5\right)-3x+15=\left(x-3\right)\left(x-5\right)\)

tt thôi nha dài lắmTruong minh quan cái này dễ mà cũng hỏi

a/ x^3-6x^2+12x-8

=(x-2)^3

b/x^2+5x+4

=x^2+x+4x+4

=x(x+1)+4(x+1)

=(x+1)(x+4)

c/ 16^2-9(x+1)^2=0

<=> (4x-3x-3)(4x+3x+3)=0

<=>x-3=0 hay 7x+3=0

<=> x=3 hay x=-3/7

d/ x^3-2x^2-x+2

=x^2(x-2)-(x-2)

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

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

e/x^2+y^2-2xy-x+y

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

f/x^3+y^3+3y^2+3y+1

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

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

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

b. x²+2.5/2x+(5/2)²-(5/2)²+4

= (x+5/2)²-25/4+4

=(x+5/2)²-(3/2)²

= x+ 5/2 -3/2 ) . (x+5/2-3/2)

= (x+1 ) (x+2)

c. 

(4x)²- [3(x+1)]² =0

[4x-3(x+1)] [4x+3(x+1)] =0

(x-3) (7x+3) =0

<=> x-3 =0 => x = 3

7x+3=0 => x= -3/7

d. x³-2x²-x+2

= (x³-2x²) - (x+2) 

= x² (x-2) - (x-2)

= (x-2) (x²-1)

CHÚC BẠN HỌC TỐT 

* Tớ còn a, e, và f sorry nó k dễ để suy nghĩ trong thơi gian ngắn được nên tớ bỏ !! Ahihihih

#Đào lê anh thư : bạn ơi câu e cậu còn thiếu thì phở ???

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

a) Ta có :\(x^3-3.x^2.2+3.x+2^2+2^3\)(Hằng đẳng thức số 5 đấy bạn)

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

b) Ta có:\(x^2+5x+4=x^2+4x+1x+4\)

\(=\left(x^2+4x+4\right)+x\)

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

c) Ta có :\(16x^2-9\left(x+1\right)^2=0\)

\(\left[4x+3\left(x+1\right)\right].\left[4x-3\left(x+1\right)\right]=0\)(Hằng đẩng thức số 3)

\(\left(4x+3x+3\right).\left(4x-3x-3\right)=0\)

 \(7x+3.\left(x-3\right)=0\)

\(\Rightarrow7x+3=0\)hoặc \(x-3=0\)

\(\Rightarrow7x=-3\)    hoặc \(x=0+3\)

\(\Rightarrow x=\frac{-3}{7}\)      hoặc \(x=3\)

Vậy:\(x=\frac{-3}{7};3\)

d) Ta có \(x^3-2x^2-x+2\)

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

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

Bây giờ hơi trễ rồi để mai mình làm tiếp 2 câu cuối nhá.

Rất vui khi được giúp bạn !!1 : =))

Phân tích đa thức thành nhân tử

c: \(x^2-4x^2y^2+y^2+2xy\)

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

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

d: \(7x^2-14xy^2+7y^4\)

\(=7\left(x^2-2xy^2+y^4\right)\)

\(=7\left(x-y^2\right)^2\)