Phân tích đa thức thành nhân tử

a)\(8x^3+\dfrac{1}{27}\)

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

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

\(=\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)

b)\(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)

\(=\left(x-y+5\right)^2-2.\left(x-y+5\right).1+1^2\)

\(=\left(x-y+5-1\right)^2\)

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

c)\(125-x^6\)

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

\(=\left(5-x^2\right)\left(5^2+5x^2+\left(x^2\right)^2\right)\)

\(=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)

d)\(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)

\(=\left(x^2+4y^2-5\right)^2-4^2\left(\left(xy\right)^2+2xy.1+1^2\right)\)

\(=\left(x^2+4y^2-5\right)^2-4^2\left(xy+1\right)^2\)

\(=\left(x^2+4y^2-5\right)^2-\left(4xy+4\right)^2\)

\(=\left(x^2+4y^2-5-4xy-4\right)\left(x^2+4y^2-5+4xy+4\right)\)

\(=\left(x^2-2.x.2y+\left(2y\right)^2-9\right)\left(x^2+2.x.2y+\left(2y\right)^2-1\right)\)

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

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

Đây bạn

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

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

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

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

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

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

8: \(x^2y-xy^2-2x+2y\)

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

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

9: \(x^2-y^2-2x+2y\)

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

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

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

tách nhỏ câu hỏi ra bạn

\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

a) 

(x-y+5)²-2.(x-y+5)+1

=(x-y+5-1)²

=(x-y+4)²

b)

(x²+4y²-5)²-16.(x².y²+2xy+1)

=(x²+4y²-5)²-[4.(xy+1)]²

=(x²+4y²-5-4xy-4)(x²+4y²-5+4xy+4)

=(x²+4y²-4xy-9)(x²+4y²+4xy-1)

=[(x-2y)²-9][(x+2y)²-1]

=(x-2y-3)(x-2y+3)(x+2y-1)(x+2y+1)

=(x²+x-3x-3)((x-2y+3)(x+2y-1)(x+1)²

=[x(x+1)-3(x+1)](x-2y+3)(x+2y-1)(x+1)²

=(x+1)(x-3)(x-2y+3)(x+2y-1)(x+1)²

a: =-4xyz^2

b: =-9x^2y

c: =16x^2y^2

d: =1/6x^2y^3

e: =13/6x^3y^2

f: =7/12x^4y

a) -xyz² - 3xz.yz

= -xyz² - 3xyz²

= -4xyz²

b) -8x²y - x.(xy)

= -8x²y - x²y

= -9x²y

c) 4xy².x - (-12x²y²)

= 4x²y² + 12x²y²

= 16x²y²

d) 1/2 x²y³ - 1/3 x²y.y²

= 1/2 x²y³ - 1/3 x²y³

= 1/6 x²y³

e) 3xy(x²y) - 5/6 x³y²

= 3x³y² - 5/6 x³y²

= 13/6 x³y²

f) 3/4 x⁴y - 1/6 xy.x³

= 3/4 x⁴y - 1/6 x⁴y

= 7/12 x⁴y

a) \(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)

\(=-\left(x-5\right)^2\left(x+5\right)^2\)

b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1=\left(x-y+5-1\right)^2=\left(x-y+4\right)^2\)

c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)

Có lẽ bạn ghi sai đề rồi nha. 

a: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)

đáp án: a là đúng