phan tich da thuc sau thanh nhan tu
x3-4x2+12x-27
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Lời giải:
$x^3-4x^2-12x+27$
$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$
$=x^2(x+3)-7x(x+3)+9(x+3)$
$=(x+3)(x^2-7x+9)$
\(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=\left(4y\right)^2-\left(2x+3\right)^2\)\(=\left[4y-\left(2x+3\right)\right]\left(4y+2x+3\right)=\left(4y-2x-3\right)\left(4y+2x+3\right)\)
A, m2-7m+12=m2-3m+12-4m=m(m-3)-4(m-3)=(m-4)(m-3)
B, 2x4-x3+27-54x=x3(2-x)-27(2-x)=(x3-27)(2-x)=(x-3)(x2+3x+9)(2-x)
a) m^2 -7m +12 = m^2 -3m -4m +12
=m(m -3)-4 (m- 3)
=(m-4)(m-3)
b) 2x^4-x^3 -54x+ 27
=(2m^4-x^3)- (54x - 27)
=x^3(2x-1)-27(2x-1)
=(x^3-27)(2x-1)
\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)
hc tốt nha !!!!!!!!!
\(x^4-6x^3+54x-81\)
\(=x^4+3x^3-9x^3+27x^2-27x^2 +81x-27x-81\)
\(=\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27+81\right)\)
\(=x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)
\(=\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)
\(=\left(x+3\right)\left(x-3\right)^3\)
a, = 2 x 2 - 6x +9x - 27
=(x-3).2x +9.(x-3)
=(x-3).(2x+9)
b, = 2x2-6xy+xy -3y2
= 2x.(x-3y) +y.(x-3y)
= (x-3y).(2x+y)
\(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)