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) \(\left(x-y\right)^3+8y^3\)

\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)

\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)

⇔ 16 x   -   1   =   16 4 x  

⇔ x - 1 = 4x ⇔  x   =   - 1 3

Chọn C

1: \(\left(x^2+1\right)^2-4x\left(1-x^2\right)\)

\(=x^4+2x^2+1-4x+4x^3\)

\(=x^4+4x^3+2x^2-4x+1\)

\(=\left(x^2+2x-1\right)^2\)

2: \(\left(x^2-8\right)^2+36\)

\(=x^4-16x^2+64+36\)

\(=x^4-16x^2+100\)

\(=x^4+20x^2+100-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2\)

\(=\left(x^2+6x+10\right)\left(x^2-6x+10\right)\)

3: \(x^4+4=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-4x^2\)

\(=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)

4: \(x^4+64=x^4+16x^2+64-16x^2\)

\(=\left(x^2+8\right)^2-16x^2\)

\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

5: \(64x^4+1=64x^4+16x^2+1-16x^2\)

\(=\left(8x^2+1\right)^2-16x^2\)

\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)

lỗi ah

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