Phân tích đa thức thành nhân tử:
64x4+y4
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64x^4+y^4
=64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2-4xy+y^2)(8x^2+4xy+y^2)
Lời giải:
a.
$=(x^2)^2+(\frac{1}{2}y^4)^2+2.x^2.\frac{1}{2}y^4-x^2y^4$
$=(x^2+\frac{1}{2}y^4)^2-(xy^2)^2$
$=(x^2+\frac{1}{2}y^4-xy^2)(x^2+\frac{1}{2}y^4+xy^2)$
b.
$=(\frac{1}{2}x^2)^2+(y^4)^2+2.\frac{1}{2}x^2.y^4-x^2y^4$
$=(\frac{1}{2}x^2+y^4)^2-(xy^2)^2$
$=(\frac{1}{2}x^2+y^4-xy^2)(\frac{1}{2}x^2+y^4+xy^2)$
c.
$=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$
$=(8x^2+y^2)^2-(4xy)^2=(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
d.
$=\frac{64x^4+y^4}{64}=\frac{1}{64}(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
c: \(64x^4+y^4\)
\(=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10x\right)\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
b) \(\left(a^2+b^2\right)^2-4a^2b^2\)
\(=\left(a^2-2ab+b^2\right)\left(a^2+2ab+b^2\right)\)
\(=\left(a-b\right)^2\cdot\left(a+b\right)^2\)
c) \(a^4-b^4=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)
\(x^4-y^4+2x^3y-2xy^3\)
\(=\left(x^2+y^2\right)\left(x^2-y^2\right)+2xy\left(x^2-y^2\right)\)
\(=\left(x^2-y^2\right)\left(x^2+y^2+2xy\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x+y\right)^2\)
\(=\left(x-y\right)\left(x+y\right)^3\)
\(x^4-y^4+2x^3y-2xy^3\\ =\left(x^2\right)^2-\left(y^2\right)^2+2xy\left(x^2-y^2\right)\\ =\left(x^2-y^2\right)\left(x^2+y^2\right)+2xy\left(x^2-y^2\right)\\ =\left(x^2-y^2\right)\left(x^2+y^2+2xy\right)\\ =\left(x-y\right)\left(x+y\right)\left(x+y\right)^2\\ =\left(x-y\right)\left(x+y\right)^3\)
a) \(6x^2-11x+3\)
\(=6x^2-9x-2x+3\)
\(=3x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(3x-1\right)\left(2x-3\right)\)
b) \(2x^2+3x-27\)
\(=2x^2-6x+9x-27\)
\(=2x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(2x+9\right)\left(x-3\right)\)
Lời giải:
a. $A=9x^2+15x+6xy+y^2+5y=(9x^2+6xy+y^2)+(15x+5y)$
$=(3x+y)^2+5(3x+y)=0^2+5.0=0$
b. $25x^2-y^4-5x+y^2=(25x^2-y^4)-(5x-y^2)=(5x-y^2)(5x+y^2)-(5x-y^2)$
$=(5x-y^2)(5x+y^2-1)$
\(64x^4+y^4=64x^4+16x^2y^2-16x^2y^2+y^4\)
\(=\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)