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Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)
1.
a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)
b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)
2.
a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)
= (9x-3y)(3x-y)
= 3(3x-y)(3x-y)
= 3(3x-y)^2
b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)
= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)
= \(\left(x^2-9\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)^2\)
Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)
\(=-2x^5-10x^4+x^3\)
b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)
\(=3x^2-5x+2\)
2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)
\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)
\(=\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(x-y\right)\)
b ) \(x^3-3x^2-9x+27\)
\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x^2-9\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)
a) \(x^3-3x^2-3x+1\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
b) \(4x^2+4x+1-y^2-16y-64\)
\(=\left(2x+1\right)^2-\left(y+8\right)^2\)
\(=\left(2x+1-y-8\right)\left(2x+1+y+8\right)\)
\(=\left(2x-7-y\right)\left(2x+9+y\right)\)
c) \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
d) \(\left(x^2+y^2-5\right)^2-4\left(x^2y^2+4xy+4\right)\)
\(=\left(x^2+y^2-4-1\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)
\(=\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]\)
\(=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)
a) Đặt t = x2
bthuc <=> t2 - 7t + 16
Từ đây ta không thể phân tích được :)
b) x3 - 2x2 + 5x - 4
= x3 - x2 - x2 + x + 4x - 4
= x2( x - 1 ) - x( x - 1 ) + 4( x - 1 )
= ( x - 1 )( x2 - x + 4 )
c) x3 - 2x2 + x - 3 ( phân tích hổng ra :)) )
d) 3x3 - 4x2 + 12x - 4 ( phân tích hổng ra p2 :)) )
e) 6x3 + x2 + x + 1
= 6x3 + 3x2 - 2x2 - x + 2x + 1
= 3x2( 2x + 1 ) - x( 2x - 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 3x2 - x + 1 )
f) 4x3 + 6x2 + 4x + 1
= 4x3 + 2x2 + 4x2 + 2x + 2x + 1
= 2x2( 2x + 1 ) + 2x( 2x + 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 2x2 + 2x + 1 )
Bài làm
a) 4x2 - 6x
= 2x( 2x - 3 )
b) 9x4y3 + 3x2y4
= 3x2y3( 3x2 + y )
c) x3 - 2x2 + 5x
= x( x2 - 2x + 5 )
d) 3x( x - 1 ) + 5( x - 1 )
= ( x - 1 )( 3x + 5 )
e) 2x2( x + 1 ) + 4( x + 1 )
= ( x + 1 )( 2x2 + 4 )
= ( x + 1 )2( x2 + 2 )
= 2( x + 1 )( x2 + 2 )
f) -3x - 6xy + 9xz
= -( 3x + 6xy - 9xz )
= -3x( 1 + 2y - 3z )
# Học tốt #
a)x^2.16-4xy+4y^2
<=>16.x^2-2x2y+(2y)^2
<=>16(x-2y)^2
b)x^5-x^4+x^3-x^2
<=>(x^5-x^4)+(x^3-x^2)
<=>x^4(x-1)+x^2(x-1)
<=>(x-1)(x^4+x^2)
c)x^5+x^3-x^2-1
<=>(x^5+x^3)-(x^2+1)
<=>x^3(x^2+1)-(x^2+1)
<=>(x^2+1)(x^3-1)
d)x^4-3x^3-x+3
<=>(x^4-3x^3)-(x-3)
<=>x^3(x-3)-(x_3)
<=>(x-3)(x^3-1)
\(a,x^2.16-4xy+4y^2\)
\(=16.x^2-4xy+4y^2\)
\(=16.\left[x^2-4xy+\left(2y\right)^2\right]\)
\(=16.\left(x-2y\right)^2\)
\(b,x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4+x^2\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
\(c,x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(d,x^4-3x^3-x+3\)
\(=x^3\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(b.x^4+4x^2-5=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(c.x^3-19x-30=x^3-25x+6x-30\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
tí nữa giải cho