bấm máy tính là ra đó :VV

b, x mũ 2 + 4x + 3 = ( x+ 1 ) ( x + 3 )

a) = x² + 3x + x + 3 = x( x + 3 ) + ( x + 3 ) = ( x + 3 )( x + 1 )

b) = x² - 3x - 5x + 15 = x( x - 3 ) - 5( x - 3 ) = ( x - 3 )( x - 5 )

c) = x² - 2x + 4x - 8 = x( x - 2 ) + 4( x - 2 ) = ( x - 2 )( x + 4 )

d) = x² - 4x + 3x - 12 = x( x - 4 ) + 3( x - 4 ) = ( x - 4 )( x + 3 )

e) = 4x² - 2x + 6x - 3 = 2x( 2x - 1 ) + 3( 2x - 1 ) = ( 2x - 1 )( 2x + 3 )

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

2, \(x^2+3x-10=x^2+5x-2x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)

3, \(x^2-x-12=x^2-4x+3x-12=x\left(x+3\right)-4\left(x+3\right)=\left(x-4\right)\left(x+3\right)\)

4, \(3x^2+4x-7=3x^2+7x-3x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(3x+7\right)\left(x-1\right)\)

5, \(4x^2-9y^2-5xy=4x^2-9xy+4xy-9y^2\)

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

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

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

Câu a : \(4x^3-5x^2+6x+9\)

\(=4x^3+3x^2-8x^2-6x+12x+9\)

\(=\left(4x^3+3x^2\right)-\left(8x^2+6x\right)+\left(12x+9\right)\)

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

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

Câu b : \(5x^3-12x^2+14x-4\)

\(=5x^3-10x^2-2x^2+10x+4x-4\)

\(=\left(5x^3-2x^2\right)-\left(10x^2-4x\right)+\left(10x-4\right)\)

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

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

Câu c : \(x^3-5x^2+2x+8\)

\(=x^3+x^2-6x^2-6x+8x+8\)

\(=\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(8x+8\right)\)

\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-6x+8\right)\)

\(=\left(x+1\right)\left[x^2-2x-4x+8\right]\)

\(=\left(x+1\right)\left[x\left(x-2\right)-4\left(x-2\right)\right]\)

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

Câu d : \(4x^3+5x^2+10x-12\)

\(=4x^3+8x^2-3x^2+16x-6x-12\)

\(=\left(4x^3-3x^2\right)+\left(8x^2-6x\right)+\left(16x-12\right)\)

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

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

bạn có thể check lại đề bài câu a được không ạ

là x^2 = x^5 á bạn

f) = x²( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )

g) = 4( x - y ) + ( x - y )² = ( x - y )( x - y + 4 )

h) = x³( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 )

i) = ( x - y )( x + y ) - 4( x + y ) = ( x + y )( x - y - 4 )

j) = ( x - y )( x² + xy + y² ) - 3( x - y ) = ( x - y )( x² + xy + y² - 3 )

Trả lời:

f, x³ - 4x² - 9x + 36 = ( x³ - 4x² ) - ( 9x - 36 ) = x² ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x² - 9 ) = ( x - 4 )( x - 3 )( x + 3 )

g, 4x - 4y + x² - 2xy + y² = ( 4x - 4y ) + ( x² - 2xy + y² ) = 4 ( x - y ) + ( x - y )² = ( x - y ) ( 4 + x - y )

h, x⁴ + x³ + x² - 1 = ( x⁴ + x³ ) + ( x² - 1 ) =  x³ ( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 ) 

i, x² - y² - 4x - 4y = ( x² - y² ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )

j, x³ - y³ - 3x + 3y = ( x³ - y³ ) - ( 3x - 3y ) = ( x - y )( x² + xy + y² ) - 3 ( x - y ) = ( x - y )( x² + xy + y² - 3 ) 

f) x³-4x²-9x+36

=x²(x-4)-9(x-4)

=(x-4)(x²-9)

=(x-4)(x-3)(x+3)

g) 4x-4y+x²-2xy+y²

=4(x-y)+(x-y)²

=(x-y)(4+x-y)

h) x⁴+x³+x²-1

=x³(x+1)+(x-1)(x+1)

=(x+1)(x³+x-1)

i) x²-y²-4x-4y

=(x-y)(x+y)-4(x+y)

=(x+y)(x-y-4)

j) x³-y³-3x+3y

=(x-y)(x²+xy+y²)-3(x-y)

=(x-y)(x²+xy+y²-3)

#H

1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)

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

2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)

a)\(x^3+4x^2+4x=0\)

\(\Leftrightarrow x\left(x^2+4x+4\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

b)\(\left(x+3\right)^2-4=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3-2=0\\x+3+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)

c)\(x^4-9x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

d)\(x^2-6x+9=81\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

e)\(x^3+6x^2+9x-4x=0\)

\(\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)

#H

x⁴ + 2x³ + 5x² + 4x - 12

= x⁴ + x³ + x³ + 6x² + x² - 2x² + 6x - 2x - 12

= ( x⁴ + x³ + 6x² ) + ( x³ + x² + 6x ) - ( 2x² + 2x + 12 )

= x²( x² + x + 6 ) + x( x² + x + 6 ) - 2( x² + x + 6 )

= ( x² + x + 6 )( x² + x - 2 )

= ( x² + x + 6 )( x² - x + 2x - 2 )

= ( x² + x + 6 )[ x( x - 1 ) + 2( x - 1 ) ]

= ( x² + x + 6 )( x - 1 )( x + 2 )

Bài 2: 

a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

=>-4x+2=0

hay x=1/2

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

=>x=1 hoặc x=-1

Phân tích đa thức thành nhân tử ?

Ta có: \(P=\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

Đặt \(x^2+4x+8=y\)

Khi đó: 

\(P=y^2+3xy+2x^2\)

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

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

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

\(P=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

\(P=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)