3(2x-3)(3x+2) - 2(x+4)(4x-3) + 9x(4-x)=0
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Bài 3:
\(f\left(x\right)=9x^3-\frac{1}{3}x+3x^2-3x+\frac{1}{3}x^2-\frac{1}{9}x^3-3x^2-9x+27+3x\)
\(f\left(x\right)=\left(9x^3-\frac{1}{9}x^3\right)-\left(\frac{1}{3}x+3x+9x-3x\right)+\left(3x^2-3x^2\right)+27\)
\(f\left(x\right)=\frac{80}{9}x^3-\frac{28}{3}x+27\)
Thay x = 3 vào đa thức, ta có:
\(f\left(3\right)=\frac{80}{9}.3^3-\frac{28}{3}.3+27\)
\(f\left(3\right)=240-28+27=239\)
Vậy đa thức trên bằng 239 tại x = 3
Thay x = -3 vào đa thức. ta có:
\(f\left(-3\right)=\frac{80}{9}.\left(-3\right)^3-\frac{28}{3}.\left(-3\right)+27\)
\(f\left(-3\right)=-240+28+27=-185\)
Bài 4: \(f\left(x\right)=2x^6+3x^2+5x^3-2x^2+4x^4-x^3+1-4x^3-x^4\)
\(f\left(x\right)=2x^6+\left(3x^2-2x^2\right)+\left(5x^3-x^3-4x^3\right)+\left(4x^4-x^4\right)\)
\(f\left(x\right)=2x^6+x^2+3x^4\)
Thay x=1 vào đa thức, ta có:
\(f\left(1\right)=2.1^6+1^2+3.1^4=2+1+3=6\)
Đa thức trên bằng 6 tại x =1
Thay x = - 1 vào đa thức, ta có:
\(f\left(-1\right)=2.\left(-1\right)^6+\left(-1\right)^2+3.\left(-1\right)^4=2+1+3=6\)
Đa thức trên có nghiệm = 0
a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28
b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10
c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x
d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1
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)
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
A = \(\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)\)
A = \(x^2-6x+9-4x^2+1=-3x^2-6x+10\)
B = \(\left(2x-3\right)^2-\left(x-1\right)\left(2x+1\right)\)
B = \(4x^2-12x+9-2x^2-x+2x+1\)
B = \(2x^2-11x+10\)
C = \(4x\left(x-3\right)^2-\left(4-2x\right)^2\)
C = \(4x\left(x^2-6x+9\right)-16+16x-4x^2\)
C = \(4x^3-24x^2+36x-16+16x-4x^2\)
C = \(4x^3-28x^2+52x-16\)
D = \(3x\left(x-1\right)\left(x-2\right)-x\left(2x-1\right)^2\)
D = \(\left(3x^2-3x\right)\left(x-2\right)-x\left(2x-1\right)^2\)
D = \(3x^3-6x^2-3x^2+6x-x\left(4x^2-4x+1\right)\)
D = \(3x^3-9x^2+6x-4x^3+4x^2-x\)
D = \(-x^3-5x^2+5x\)
a) (x+2)2+x(x-4)
=x2+4x+4+x2-4x
=2x2+4
b)(x-3)2-(x+3)(x-4)
=x2-6x+9-x2+4x-3x+12
=-5x+12
c) (3x+1)2+3x(2-4x)
=9x2+6x+1+6x-12x2
=-3x2+12x+1
d) (2x-4y)2-(2x-3)(2x-3y)
=4x2-16xy+16y2-4x2+6xy+6x-9y
=16y2-10xy+6x-9y
Bài 1.
a) -2x( -3x + 2 ) - ( x + 2 )2
= 6x2 - 4x - ( x2 + 4x + 4 )
= 6x2 - 4x - x2 - 4x - 4
= 5x2 - 8x - 4
b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2
= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1
= 4
d) x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
Bài 2.
a) 4x2 - 4xy + y2 = ( 2x - y )2
b) 9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )
c) x3 + 2 + 3( x3 - 2 )
= x3 + 2 + 3x3 - 6
= 4x3 - 4
= 4( x3 - 1 )
= 4( x - 1 )( x2 + x + 1 )
Bài 3.
2( x - 2 ) = x2 - 4x + 4
⇔ ( x - 2 )2 - 2( x - 2 ) = 0
⇔ ( x - 2 )( x - 2 - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ x = 2 hoặc x = 4