Phân tích đa thức thành nhân tử:
a) \(x^4+6x^3+7x^2-6x-8\)
b) \(x^2-25+14xy+4y^2\)
c) \(2x^3+5x^2-7x\)
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\(a,=2x^2-6x+9x-27=\left(x-3\right)\left(2x+9\right)\\ b,=x^2-7x+\dfrac{49}{4}-\dfrac{73}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{73}{4}=\left(x-\dfrac{7}{2}-\dfrac{\sqrt{73}}{2}\right)\left(x-\dfrac{7}{2}+\dfrac{\sqrt{73}}{2}\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ d,=x^2-2x-8x+16=\left(x-2\right)\left(x-8\right)\\ e,=x^2-3x-5x+15=\left(x-3\right)\left(x-5\right)\\ g,=x^2+2x+4x+8=\left(x+2\right)\left(x+4\right)\)
Lời giải:
a.
$x^2-7x+6=(x^2-x)-(6x-6)=x(x-1)-6(x-1)=(x-1)(x-6)$
b.
$x-3\sqrt{3}x-12\sqrt{3}$ không phân tích được thành nhân tử
c.
$x^2+4x-2$ không phân tích được thành nhân tử với các hệ số nguyên.
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5+y\right)\left(x+5-y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(x^2+2\right)\left(5x-7\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5-y\right)\left(x+5+y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(5x-7\right)\left(x^2+2\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
6x3 + 5x2 - 7x - 4
= (6x3 + 5x - 7x) - 4
= x (6x2 - 5 - 7) - 22
= x (6x2 - 12) - 22
= x [6 (x2 - 2)] - 22
= x [6 (x2 - \(\sqrt{2}^2\))] - 22
= x [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))] - 22
= (x - 22) [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))
b) 2x3 - x2 + x - 2
= (2x3 - x2 - x) - 2
= x (2x2 - x - 1) - 2
= (x - 2) (2x2 - x - 1)
(mik ko biet dug ko, neu sai mog bn thog cam)
\(a,=5x^2-5x+3x-3=\left(x-1\right)\left(5x+3\right)\\ b,=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\\ c,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ d,=7x^2-7x+x-1=\left(x-1\right)\left(7x+1\right)\)
a, x^2 + 5x +4
= x^2 + 1x + 4x + 4
= (x^2 + 1x) + (4x + 4)
= x ( x + 1 ) + 4 ( x + 1 )
= (x + 1) (x + 4)
b, x^2 - 6x + 5
= x^2 - 1x - 5x + 5
= (x^2 - 1x) - (5x - 5)
= x (x - 1) - 5 (x - 1)
= (x - 1) (x - 5)
c, x^2 + 7x + 12
= x^2 + 3x + 4x + 12
= (x^2 + 3x) + (4x + 12)
= x (x + 3) + 4 (x + 3)
= (x + 3) (x + 4)
d, 2x^2 - 5x + 3
= 2^x2 - 2x - 3x + 3
= 2x (x - 1) - 3 (x - 1)
= (x-1) (2x - 3)
e, 7x - 3x^2 - 4
= 3x + 4x - 3x^2 - 4
= (3x - 3x^2) + (4x - 4)
= 3x (1 - x) + 4 (x - 1)
= 3x (1-x) - 4 (1 - x)
= (1 - x) (3x - 4)
f, x^2 - 10x + 16
= x^2 - 2x - 8x + 16
= (x^2 - 2x) - (8x - 16)
= x (x - 2) - 8 (x - 2)
= (x - 2) (x - 8)
a, (x+1)(x+4)
b,(x-5)(x-1)
c,(x+3)(x+4)
d,(2x-3)(x-1)
e,(-3x+4)(x-1)
f, (x-8)(x-2)
\(a,=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)