a) Ta có: \(4x^4+81\)
\(=\left(2x^2\right)^2+9^2+2\cdot2x^2\cdot9-36x^2\)
\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
b) Ta có: \(64x^4+y^4\)
\(=\left(8x^2\right)^2+\left(y^2\right)^2+2\cdot8x^2\cdot y^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
c) Ta có: \(\left(2x^2-4\right)^2+9\)
\(=4x^4-16x^2+16+9\)
\(=4x^4-16x^2+25\)
\(=4x^4+20x^2+25-36x^2\)
\(=\left(2x^2+5\right)^2-\left(6x\right)^2\)
\(=\left(2x^2-6x+5\right)\left(2x^2+6x+5\right)\)
Lời giải:
a) $4x^4+81=(2x^2)^2+9^2=(2x^2)^2+9^2+2.2x^2.9-36x^2$
$=(2x^2+9)^2-(6x)^2=(2x^2+9-6x)(2x^2+9+6x)$
b)
$64x^4+y^4=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$
$=(8x^2+y^2)^2-(4xy)^2=(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
c) Biểu thức không phân tích được thành nhân tử.
