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Lời giải:
a.
$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$
b.
$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$
c.
$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$
d.
$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$
$=(x+y)(x^2-4xy+7y^2)$
a) \(64x^2-24y^2\)
\(=8\left(8x^2-3y^2\right)\)
b) \(64x^3-27y^3\)
\(=\left(4x\right)^3-\left(3y\right)^3\)
\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c) \(x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
d) \(\left(x-y\right)^3+8y^3\)
\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)
\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)
x^3-64x=0
<=>x^3-64x=(x-8)x(x+8)
=>(x-8)x(x+8)=0
Th1:x-8=0
=>x=8
Th2:x+8=0
=>x=-8
vậy pt có x=±8
x^3-64x=0
x.x.x-64x=0
=>x.x=64
Ta có:
x.x=64
x=\(\sqrt{64}\)
x=8
\(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)
\(\Leftrightarrow64x^3=\left(x-2+3x+2\right)\left(\left(x-2\right)^2-\left(x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\right)\)\(\Leftrightarrow16x^2=\left(x-2\right)^2-\left(x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\)
\(\Leftrightarrow x^2-4x+4-3x^2-2x+6x+4+9x^2+12x+4-16x^2=0\)
\(\Leftrightarrow-9x^2+12x+12\Leftrightarrow3x^2-4x-4=0\).Dùng Casio bấm nghiệm nhá! mk mất máy tính rồi!!!!
\(x^3-64x\)
\(=x\left(x^2-64\right)=x\left(x^2-8^2\right)\)
\(=x\left(x-8\right)\left(x+8\right)\)
1 ) ...
2 ) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}\right)^2+2.\dfrac{x}{2}.2y+\left(2y\right)^2=\left(\dfrac{x}{2}+2y\right)^2\)
3 ) \(64x^3-1=\left(4x\right)^3-1=\left(4x-1\right)\left[\left(4x\right)^2+4x+1\right]=\left(4x-1\right)\left(16x^2+4x+1\right)\)
4 ) \(25-10y+y^2=5^2-2.5y+y^2=\left(5-y\right)^2\)
a: =64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2-4xy)(8x^2+y^2+4xy)
b: =x^8+2x^4+1-x^4
=(x^4+1)^2-x^4
=(x^4-x^2+1)(x^4+x^2+1)
=(x^4-x^2+1)(x^4+2x^2+1-x^2)
=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)
c: =(x+1)(x^2-x+1)+2x(x+1)
=(x+1)(x^2-x+1+2x)
=(x+1)(x^2+x+1)
d: =(x^2-1)(x^2+1)-2x(x^2-1)
=(x^2-1)(x^2-2x+1)
=(x-1)^2*(x-1)(x+1)
=(x+1)(x-1)^3