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a) x2 - 5x + 6
= x2 - 2x - 3x + 6
=(x2 - 2x) - (3x + 6)
=x.(x - 2) - 3.(x - 2)
=(x-2).(x-3)
b) 3x2+9x-30
=3x2+15x-6x-30
=(3x2+15x) - (6x+30)
= 3x(x+5) - 6(x+5)
=(x+5).(3x-6)
c) x2-3x+2
=x2-2x-x+2
=(x2-2x) - (x-2)
=x(x-2)-(x-2)
=(x-2)(x-1)
a)x2 - 5x + 6
= x2 - 2x - 3x + 6
=x.(x - 2) - 3(x - 2)
=(x - 2).(x - 3)
b)3x2 +9x -30
=3x2 +15x - 6x -30
=3x.(x+5) - 6.(x + 5)
=(x+5).(3x - 6)
c)x2 - 3x +2
=x2 - 2x - x +2
=x.(x- 2) - 1.(x-2)
=(x-2).(x - 1)
d)x2 - 9x +18
=x2 - 6x -3x +18
=x.(x - 6) -3.(x - 6)
=(x - 6).(x - 3)
e)x2 - 6x +8
=x2 - 2x - 4x +8
=x.(x - 2)- 4.(x - 2)
=(x - 2).(x - 4)
f)x2 - 5x -14
=x2 + 2x - 7x - 14
=x.(x + 2) -7.(x + 2)
=(x + 2).(x - 7)
a) \(x^2\)\(-5x+6\)
=\(x^2\)\(-3x-2x+6\)
=\(x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-2\right)\left(x-3\right)\)
b) \(3x^2\)\(+9x-30\)
=\(3x^2\)\(-6x+15x-30\)
=\(3x\left(x-2\right)+15\left(x-2\right)\)
=\(\left(x-2\right)\left(3x+15\right)\)
c)\(x^2\)\(-3x+2\)
=\(x^2\)\(-2x-x+2\)
=\(x\left(x-2\right)-\left(x-2\right)\)
=\(\left(x-2\right)\left(x-1\right)\)
d) \(12x^2\)\(+7x-12\)
=\(12x^2\)\(-9x+16x-12\)
=\(3x\left(4x-3\right)+4\left(4x-3\right)\)
=\(\left(3x+4\right)\left(4x-3\right)\)
e) \(15x^2\)\(+7x-2\)
=\(15x^2\)\(-3x+10x-2\)
=\(3x\left(5x-1\right)+2\left(5x-1\right)\)
=\(\left(3x+2\right)\left(5x-1\right)\)
f) \(a^2\)\(-5a-14\)
=\(a^2\)\(-7a+2a-14\)
=\(a\left(a-7\right)+2\left(a-7\right)\)
=\(\left(a+2\right)\left(a-7\right)\)
g) \(x^2\)\(-\left(a+b\right)x+ab\)
=\(x^2\)\(-ax-bx+ab\)
=\(x\left(x-a\right)-b\left(x-a\right)\)
=\(\left(x-a\right)\left(x-b\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) x2 + x - 12 = x2 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )
b) x2 - 4x - 5 = x2 + x - 5x - 5 = x( x + 1 ) - 5( x + 1 ) = ( x + 1 )( x - 5 )
c) x2 - 2x - 3 = x2 + x - 3x - 3 = x( x + 1 ) - 3( x + 1 ) = ( x + 1 )( x - 3 )
d) x2 - 2x - 8 = x2 + 2x - 4x - 8 = x( x + 2 ) - 4( x + 2 ) = ( x + 2 )( x - 4 )
e) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
f) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x( x - 2 ) - 4( x - 2 ) = ( x - 2 )( x - 4 )
g) x2 + 4x + 3 = x2 + x + 3x + 3 = x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( x + 3 )
h) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )
i) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
j) x2 - 5x - 14 = x2 + 2x - 7x - 14 = x( x + 2 ) - 7( x + 2 ) = ( x + 2 )( x - 7 )
a) x2 + 5x + 4
= x2 + x + 4x + 4
= x (x+1) + 4 (x+1)
= (x+1) ( x+4)
c) x2 - 7x + 12
= x2 - 3x - 4x +12
= x(x-3) - 4(x-3)
= (x-3)( x-4)
m) \(5x^2+6x+1\)
\(=5x^2+5x+x+1\)
\(=5x\left(x+1\right)+\left(x+1\right)\)
\(=\left(5x+1\right)\left(x+1\right)\)
a,x2 - 13x + 36
= x2 - 4x - 9x + 36
= x(x - 4) - 9(x - 4)
= (x - 4)(x - 9)
b,x2 + 3x - 18
= x2 - 9 + 3x - 9
= (x - 3)(x + 3) + 3(x - 3)
= (x - 3)(x + 3 + 3)
= (x - 3)(x + 6)
c,x2 - 5x - 24
= x2 + 3x - 8x - 24
= x(x + 3) - 8(x + 3)
= (x + 3)(x - 8)
d,3x2 - 16x + 5
= 3x2 - 15x - x + 5
= 3x(x - 5) - (x - 5)
= (x - 5)(3x - 1)
e, 8x2 + 30x + 7
= 8x2 + 2x + 28x + 7
= 2x(4x + 1) + 7(4x + 1)
= (4x + 1)(2x + 7)
g,2x2 - 5x - 12
= 2x2 - 8x + 3x - 12
= 2x(x - 4) + 3(x - 4)
= (x - 4)(2x + 3)
i,x4 + 4
= x4 + 4x2 + 4 - 4x2
= (x2 + 2)2 - (2x)2
= (x2 + 2 - 2x)(x2 + 2 + 2x)
Phân tích đa thức thành nhân tử bằng cách tách một số hạng tử thành nhiều số hạng khác
a) x^2+4x+3
b) 4x^2-4x-3
c)x^2-x-12
d) 4x^4-4x^2-8y^4
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
1. a)\(x^2+x-3x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
a/\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)b/
\(3x^2+9x-30=3\left(x^2+3x-10\right)\)
c/
\(x^2-3x+2=x^2-x-2x+2=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
d/\(x^2-9x+18=x^2-3x-6x+18=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)e/
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)f/\(x^2-5x-14=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x+2\right)\left(x-7\right)\)
g/
\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
h/
\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-4\right)\left(x-3\right)\)i/\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) Ta có: \(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
b) Ta có: \(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2+5x-2x-10\right)\)
\(=3\left[x\left(x+5\right)-2\left(x+5\right)\right]\)
\(=3\left(x+5\right)\left(x-2\right)\)
c) Ta có: \(x^2-3x+2\)
\(=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
d) Ta có: \(x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
e) Ta có: \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
f) Ta có: \(x^2-5x-14\)
\(=x^2-7x+2x-14\)
\(=x\left(x-7\right)+2\left(x-7\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
g) Ta có: \(x^2-6x+5\)
\(=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(x-5\right)\)
h) Ta có: \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
i) Ta có: \(x^2-7x+10\)
\(=x^2-2x-5x+10\)
\(=x\left(x-2\right)-5\left(x-2\right)\)
\(=\left(x-2\right)\left(x-5\right)\)