x²-9xy+20y²=x²-4xy-5xy+20y²

nhóm 2 cái đầu, 2 cái cuối.

1.  =2(x²-2xy+y^2-16)

     =2((x-y)²-4²)

     =2(x-y-4)(x-y+4)

2.   =5(x²-2.1/10.x+1/100-y²+2.1/10y-1/100)

      =5.((X-1/10)²-(Y-1/10)²)

      =5(X-1/10-Y+1/10)(X-1/10+Y-1/10)

      =5(X-Y)(X+Y-2/10)

3.   =(X²-9XY+81/4Y²)-1/4Y²

      =(X-9/2Y)^2-1/4Y^2

     =(X-9/2Y-1/2Y)(X-9/2Y+1/2Y)

     (X-5Y)(X-4Y)

(1+x2)2−4x(1−x2)

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

đặt \(\left(1-x^2\right)\)= a

ta có :

- a . a - 4x .a

= a ( - a - 4x )

thay a = \(\left(1+x^2\right)\) ta có

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

phân tích tiếp nhé !
 

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

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

\(=\left(4-x-4\right)\left(4-x+4\right)\)

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

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

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

Phân tích các đa thức sau thành nhân tử ... c) 6x(x+y)^2+3x^2y(x+y). 2: .... x³ - 5x + 8x - 4=x² . x -5x + 8x -2² = (x² - 2²) . (x -5x + 8x )=(x-2) . (x+2) . 4x. x³ - 9x² ..... Phân tích các đa thức sau thành nhân tử : a,x^3+5x^2+8x+4 b, x^3-9x^2+6x+16 .

\(a.=x^2\left(x-y\right)-25\left(x-y\right)\)

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

\(b.=\left(b-a\right)\left(b+a\right)-4\left(b-a\right)\)

\(=\left(b+a-4\right)\left(b-a\right)\)

a ) x³ - x²y - 25x + 25y

= x² . ( x - y ) - 25 . ( x - y )

= ( x - y )( x² - 25 )

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

b ) b² - 4b + 4a - a²

= ( b² - a² ) - ( 4b - 4a )

= ( b - a )( a + b ) - 4 . ( b - a )

= ( b - a )( a + b - 4 )

c ) x³ - 4x² - 8x + 8

= ( x³ + 8 ) - ( 4x² + 8x )

= ( x + 2 )( x² - 2x + 4 ) - 4x . ( x + 2 )

= ( x + 2 )( x² - 2x + 4 - 4x )

= ( x + 2 )( x² - 6x + 4 )

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: \(x^2-y^2-2x+2y\)

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

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

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

4x(x+y)(x+y+z)(x+z)+y²z²=4(x²+xy+xz)(x²+xy+xz+yz)+y²z²=4(x²+xy+xz)²+4yz(x²+xy+xz)+y²z²=(2(x²+xy+xz)+yz)²=(2x²+2xy+2xz+yz)