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b) \(3x^2+2x-5=3\left(x-1\right)\left(x+\dfrac{5}{3}\right)\)
c) \(3-2x-x^2=-\left(x-1\right)\left(x+3\right)\)
d) \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
e) \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
b: \(3x^2+2x-5\)
\(=3x^2-3x+5x-5\)
\(=\left(x-1\right)\left(3x+5\right)\)
c: \(3-2x-x^2\)
\(=-\left(x^2+2x-3\right)\)
\(=-\left(x+3\right)\left(x-1\right)\)
d: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
e: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
`a. =4(x^2+4x+3)=4(x^2+3x+x+3)=4(x+3)(x+1)`
`b. =x^2+8x-7x-56=x(x+8)-7(x+8)=(x+8)(x-7)`
`c. =x^2-9x+8x-72=x(x-9)+8(x-9)=(x-9)(x+8)`
`d. =(x-y)^2-9=(x-y-3)(x-y+3)`
a) ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 15
= [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 15
= ( x2 + 5x + 4 )( x2 + 5x + 6 ) - 15 (*)
Đặt t = x2 + 5x + 4
(*) trở thành
t( t + 2 ) - 15
= t2 + 2t - 15
= t2 - 3t + 5t - 15
= t( t - 3 ) + 5( t - 3 )
= ( t - 3 )( t + 5 )
= ( x2 + 5x + 4 - 3 )( x2 + 5x + 4 + 5 )
= ( x2 + 5x + 1 )( x2 + 5x + 9 )
b) ( x + 2 )( x + 3 )2( x + 4 ) - 12
= [ ( x + 2 )( x + 4 ) ]( x + 3 )2 - 12
= ( x2 + 6x + 8 )( x2 + 6x + 9 ) - 12 (*)
Đặt t = x2 + 6x + 8
(*) trở thành
t( t + 1 ) - 12
= t2 + t - 12
= t2 - 3t + 4t - 12
= t( t - 3 ) + 4( t - 3 )
= ( t - 3 )( t + 4 )
= ( x2 + 6x + 8 - 3 )( x2 + 6x + 8 + 4 )
= ( x2 + 6x + 5 )( x2 + 6x + 12 )
= ( x2 + x + 5x + 5 )( x2 + 6x + 12 )
= [ x( x + 1 ) + 5( x + 1 ) ]( x2 + 6x + 12 )
= ( x + 1 )( x + 5 )( x2 + 6x + 12 )
a, Gọi\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt\(y=x^2+5x+4\)
\(\Rightarrow A=y\left(y+2\right)-15\)
\(=y^2+2y-15\)
\(=\left(x-3\right)\left(x+5\right)\)
Hay\(A=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
Vậy...
b,Gọi\(B=\left(x+2\right)\left(x+3\right)^2\left(x+4\right)-12\)
\(=\left(x^2+6x+8\right)\left(x^2+6x+9\right)-12\)
Đặt\(z=x^2+6x+8\)
\(\Rightarrow B=z\left(z+1\right)-12\)
\(=z^2+z-12\)
\(=\left(z-3\right)\left(z+4\right)\)
Hay\(B=\left(x^2+6x+5\right)\left(x^2+6x+12\right)\)
Vậy...
Linz
Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
\(a,\left(x^2+x+1\right)\left(x^2+x+2\right)-12.\)
Đặt \(x^2+x+1=a\)
\(\Rightarrow a\left(a+1\right)-12\)\(=a^2+a-12\)
\(=a^2-3a+4a-12\)
\(=a\left(a-3\right)+4\left(a-3\right)\)
\(=\left(a-3\right)\left(a+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(b,\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=a\)
\(\Rightarrow a^2+4a-12\)
\(=a^2-2a+6a-12\)
\(=a\left(a-2\right)+6\left(a-2\right)\)
\(=\left(a-2\right)\left(a+6\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
a) 5x^2 + 6xy + y^2
=5x2+5xy+xy+y2
=5x.(x+y)+y.(x+y)
=(x+y)(5x+y)
b) x^2 + 2xy - 15y^2.
=x2-3xy+5xy-15y2
=x.(x-3y)+5y.(x-3y)
=(x-3y)(x+5y)
c) (x-y)^2 + 4(x-y) - 12
=(x-y)2+4(x-y)+4-16
=(x-y+2)2-16
=(x-y+2-4)(x-y+2+4)
=(x-y-2)(x-y+6)
d) x^3 - 2x - 4.
=x3+2x2+2x-2x2-4x-4
=x.(x2+2x+2)-2.(x2+2x+2)
=(x2+2x+2)(x-2)
\(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)-4x^2\)
\(=\left(x+2\right)\left(x+12\right)\left(x+3\right)\left(x+8\right)-4x^2\)
\(=\left(x^2+14x+24\right)\left(x^2+11x+24\right)-\left(2x\right)^2\)
Đặt \(x^2+11x+24=a\)
\(=a\left(a+3x\right)-4x^2=a^2+3ax-4x^2=a^2-ax+4ax-4x^2=\left(a-x\right)\left(a+4x\right)\)
Ta có x 3 + 12 x = x . x 2 + x . 12 = x ( x 2 + 12 )
Đáp án cần chọn là: B