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a) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
\(a)\)
\(4x^2-y^2+2x+y\)
\(=\left(4x^2-y^2\right)+\left(2x+y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)\)
\(=\left(2x+y\right)\left(2x-y+1\right)\)
\(b)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x-9\right)\)
\(=\left(x-3\right)\left(x^2+5-9\right)\)
\(c)\)
\(12x^3+4x^2-27x-9\)
\(=\left(12x^3+4x^2\right)-\left(27x+9\right)\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)[\left(2x\right)^2-3^2]\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(d)\)
\(16x^2+4x-y^2+y^2\)
\(=16x^2+4x\)
\(4x\left(4x+1\right)\)
a) 1 - 2y + y2
= (1-y)2
b) ( x + 1 )2 - 25
=( x + 1 )2 - 52
=(x+1+5)(x+1-5)
b)3x^2-18x+27=3x^2-9x-9x+27=3x*(x-3)-9*(x-3)=(x-3)*(3x-9)=(x-3)*3*(x-3)=3*(x-3)^2
c)x^3-4x^2-12x+27=(x+3)*(x^2-3x+9-4)=(x+3)*(x^2-3x+5)
d)27x^3-1/27=(3x-1/3)*(9x^2-x+1/9) (hang dt)
con a) voi e) mk chiu
a) x3 - x2 - 5x + 125
=(x3-6x2+25x)+(5x2-30x+125)
=x(x2-6x+25)+5(x2-6x+25)
=(x+5)(x2-6x+25)
b) x3 + 2x2 - 6x - 27
=x3+5x2+9-3x2-15x-27
=x(x2+5x+9)-3(x2+5x+9)
=(x-3)(x2+5x+9)
c) 12x3 + 4x2 - 27x - 9
=4x2(3x+1)-9(3x+1)
=(4x2-9)(3x+1)
=[(2x)2-32](3x+1)
=(2x-3)(2x+3)(3x+1)
a) \(x^3-x^2-5x+125\)
\(=\left(x^3+125\right)-\left(x^2+5x\right)\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(x^3+2x^2-6x-27\)
\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+9+2x\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
c) \(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x-1\right)\left(4x^2-9\right)=\left(3x-1\right)\left(2x-3\right)\left(2x+3\right)\)
a) \(4x^2-12x+9=\left(2x\right)^2-2\cdot2x\cdot3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2\cdot1\cdot6x+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2\cdot3x\cdot4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
e) \(8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2+2x+1\right)\)
f) \(-8x^3+27=3^3-\left(2x\right)^3=\left(3-2x\right)\left(9+6x+4x^2\right)\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
\(1-2y+y^2=\left(1-y\right)^2\)
\(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
\(27+27x+9x^2=9\left(3+3x+x^2\right)\)
\(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
\(3x^2-6xy+9y^2=3\left(x^2-2xy+3y^2\right)\)
\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
\(x^2-4x-5=x^2+x-5x-5=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
a ) \(1-2y+y^2=y^2-2y+1=\left(y-1\right)^2\)
b ) \(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right).\)
c ) \(1-4x^2=\left(1-2x\right)\left(1+2x\right).\)
d ) \(27+27x+9x^2=9\left(3+3x+x\right)=9\left(3+4x\right).\)
e ) \(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
f ) \(3x^2-6xy+9y^2=3\left(x^2-2xy+3y^2\right).\)
g ) \(x^2+4x+3==x^2+3x+x+3=\left(x+1\right)\left(x+3\right)\)
h ) \(x^2-4x-5=x^2+x-5x-5=\left(x-5\right)\left(x+1\right).\)
a, 4x3 -12x2 + 9x
=x(4x2 -12x + 9)
=x((2x)2 - 2.3.2x + 32)
=x(2x - 3)2
b,ab + c2 -ac - bc
=(ab - ac) + (c2 - bc)
=a(b - c) + c(c - b)
=a(b - c) - c(b - c)
=(a - c)(b - c)
c,4x2 - y2 + 1 - 4x
=((2x)2 - 2.2x + 1) - y2
=(2x - 1)2 - y2
=(2x - y -1)(2x + y - 1)
d,6x2 - 7x - 20
= -(-6x2 + 7x + 20)
= -(-6x2 + 11x +10 + 10 - 4x)
= -((3x + 2)(-2x + 5) + 10 - 4x)
= -(3x + 2)(-2x + 5) -10 + 4x
= -(3x + 2)(-2x + 5) - 2(-2x + 5)
= -(-2x + 5)(3x + 4)
a) Ta có: \(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1\)
\(=\left(2x+1\right)^2\)
b) Sửa đề: \(x^2-6x+9-9y^2\)
Ta có: \(x^2-6x+9-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
c) Ta có: \(12x-9-4x^2\)
\(=-\left(4x^2-12x+9\right)\)
\(=-\left[\left(2x\right)^2-2\cdot2x\cdot3+3^2\right]\)
\(=-\left(2x-3\right)^2\)
d) Ta có: \(1-9x+27x^2-27x^3\)
\(=1^3-3\cdot1^2\cdot3x+3\cdot1\cdot\left(3x\right)^2-\left(3x\right)^3\)
\(=\left(1-3x\right)^3\)