Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{x^4}{y^4} + 4 = \left( {{x^4}{y^4} + 4{x^2}{y^2} + 4} \right) - 4{x^2}{y^2}\\
 = {\left( {{x^2}{y^2} + 2} \right)^2} - {\left( {2xy} \right)^2}\\
 = \left( {{x^2}{y^2} + 2 - 2xy} \right)\left( {{x^2}{y^2} + 2 + 2xy} \right)\\
b,\\
4{x^4} + 1 = \left( {4{x^4} + 4{x^2} + 1} \right) - 4{x^2}\\
 = {\left( {2{x^2} + 1} \right)^2} - {\left( {2x} \right)^2}\\
 = \left( {2{x^2} + 1 - 2x} \right)\left( {2{x^2} + 1 + 2x} \right)\\
c,\\
64{x^4} + 1 = \left( {64{x^4} + 16{x^2} + 1} \right) - 16{x^2}\\
 = {\left( {8{x^2} + 1} \right)^2} - {\left( {4x} \right)^2}\\
 = \left( {8{x^2} + 1 - 4x} \right)\left( {8{x^2} + 1 + 4x} \right)\\
d,\\
{x^4} + 64\\
 = \left( {{x^4} + 16{x^2} + 64} \right) - 16{x^2}\\
 = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2}\\
 = \left( {{x^2} + 8 - 4x} \right)\left( {{x^2} + 8 + 4x} \right)\\
e,\\
16{x^4}\left( {x - y} \right) - x + y\\
 = 16{x^4}\left( {x - y} \right) - \left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {16{x^4} - 1} \right)\\
 = \left( {x - y} \right)\left( {4{x^2} - 1} \right)\left( {4{x^2} + 1} \right)\\
 = \left( {x - y} \right)\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\\
f,\\
2{x^3}y - 2x{y^3} - 4x{y^2} - 2xy\\
 = 2xy\left( {{x^2} - {y^2} - 2y - 1} \right)\\
 = 2xy.\left[ {{x^2} - \left( {{y^2} + 2y + 1} \right)} \right]\\
 = 2xy.\left[ {{x^2} - {{\left( {y + 1} \right)}^2}} \right]\\
 = 2xy\left( {x + y + 1} \right)\left( {x - y - 1} \right)\\
g,\\
x\left( {{y^2} - {z^2}} \right) + y\left( {{z^2} - {x^2}} \right) + z\left( {{x^2} - {y^2}} \right)\\
 = x\left( {y - z} \right)\left( {y + z} \right) + y{z^2} - y{x^2} + z{x^2} - z{y^2}\\
 = x\left( {y - z} \right)\left( {y + z} \right) + yz\left( {z - y} \right) - {x^2}\left( {y - z} \right)\\
 = \left( {y - z} \right)\left( {xy + xz - yz - {x^2}} \right)\\
 = \left( {y - z} \right)\left[ {y\left( {x - z} \right) + x\left( {z - x} \right)} \right]\\
 = \left( {y - z} \right)\left( {x - z} \right)\left( {y - x} \right)\\
h,\\
16{x^3} - 54{y^3}\\
 = 2\left( {8{x^3} - 27{y^3}} \right)\\
 = 2.\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right)\\
i,\\
5{x^2} - 5{y^2} = 5\left( {{x^2} - {y^2}} \right) = 5\left( {x - y} \right)\left( {x + y} \right)\\
k,\\
16{x^3}y + \frac{1}{4}y{z^3}\\
 = \frac{1}{4}y\left( {64{x^3} + {z^3}} \right)\\
 = \frac{1}{4}y.\left( {4x + z} \right)\left( {16{x^2} - 4xz + {z^2}} \right)\\
l,\\
2{x^4} - 32 = 2.\left( {{x^4} - 16} \right) = 2.\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right) = 2\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)
\end{array}\)

Đáp án:

`a) x^4y^4+4`
`=x^4y^4+4x^2y^2+4-4x^2y^2`
`=(x^4y^4+4x^2y^2+4)-4x^2y^2`
`=(x^2y^2+2)^2-4x^2y^2`
`=(x^2y^2+2-2xy)(x^2y^2+2+2xy)`
`b) 4x^4+1`
`=4x^4+4x^2+1-4x^2`
`=(4x^4+4x^2+1)-4x^2`
`=(2x^2+1)-4x^2`
`=(2x^2+1-2x)(2x^2+1+2x)`
`c) 64x^4+1`
`=64x^4+16x^2+1-16x^2`
`=(64x^4+16x^2+1)-16x^2`
`=(8x^2+1)^2-8x^2`
`=(8x^2+1+4x)(8x^2+1-4x)`
`d) x^4+64`
`=x^4+16x^2+64-16x^2`
`=(x^4+16x^2+64)-16x^2`
`=(x^2+8)^2-16x^2`
`=(x^2+8+4x)(x^2+8-4x)`
`e) 16x^4(x-y)-x+y`
`=16x^4(x-y)-(x-y)`
`=(x-y)(16x^4-1)`
`=(x-y)(4x^2+1)(4x^2-1)`
`f) 2x^3y-2xy^3-4xy^2-2xy`
`=2xy(x^2-y^2-2y-1)`
`=2xy[x^2-(y^2+2y+1)]`
`=2xy[x^2-(y+1)^2]`
`=2xy(x+y+1)[x-(y+1)]`
`=2xy(x+y+1)(x-y-1)`
`g)``x(y^2- z^2)+y(z^2- x^2)+ z(x^2- y^2)`
`=``x(y^2- z^2)+y(y^2 + z^2- x^2- y^2)+z(x^2 -y^2)`
`=x(y^2-z^2)-y(y^2-z^2)-y(x^2-y^2)+z(x^2-y^2)``
$=[x(y^2-z^2)-y(y^2-z^2)]+[-y(x^2-y^2)+z(x^2-y^2)]$
`=(x-y)(y^2-z^2)-(y-z)(x^2-y^2)`
`=(x-y)(y-z)(y+z)-(y-z)(x-y)(x+y)`
`=(x-y)(y-z)[y+z-(x+y)]`
`=(x-y)(y-z)(y+z-x-y)`
`=(x-y)(y-z)(z-x)`
`h) 16x^3-54y^3`
`=2.(8x^3-27y^3)`
`=2.(2x-3y)(4x^2+6xy+9y^2)`
`i) 5x^2-5y^2`
`=5(x^2-y^2)`
`=5(x-y)(x+y)`
`k) 16x^3y+1/4 yz^3`
`=2y(8x^3+1/8 z^3)`
`=2y(2x+1/2z)(4x^2+xz+1/4 z^2)`
`l) 2x^4-32`
`=2.(x^4-16)`
`=2.(x^2 -4)(x^2 +4)`
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