Đáp án:

$\begin{array}{l}
a)2{x^2} + 3x - 5\\
 = 2{x^2} + 5x - 2x - 5\\
 = \left( {2x + 5} \right)\left( {x - 1} \right)\\
b){x^2} + 4x - {y^2} + 4\\
 = {x^2} + 4x + 4 - {y^2}\\
 = {\left( {x + 2} \right)^2} - {y^2}\\
 = \left( {x + 2 + y} \right)\left( {x + 2 - y} \right)\\
c)2{x^2} - 18\\
 = 2\left( {{x^2} - 9} \right)\\
 = 2\left( {x - 3} \right)\left( {x + 3} \right)\\
d){x^4}{y^4} + 4\\
 = {x^4}{y^4} + 4{x^2}{y^2} + 4 - 4{x^2}{y^2}\\
 = {\left( {{x^2}{y^2} + 2} \right)^2} - {\left( {2xy} \right)^2}\\
 = \left( {{x^2}{y^2} + 2 + 2xy} \right)\left( {{x^2}{y^2} + 2 - 2xy} \right)\\
e){x^7} + {x^2} + 1\\
 = {x^7} + {x^6} + {x^5} - {x^6} - {x^5} - {x^4}\\
 + {x^4} + {x^3} + {x^2} - {x^3} - {x^2} - x + {x^2} + x + 1\\
 = {x^5}\left( {{x^2} + x + 1} \right) - {x^4}\left( {{x^2} + x + 1} \right)\\
 + {x^2}\left( {{x^2} + x + 1} \right) - x\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
 = \left( {{x^2} + x + 1} \right)\left( {{x^5} - {x^4} + {x^2} - x + 1} \right)
\end{array}$

$\\$

`a,`

`2x^2 +3x-5`

`= 2x^2 - 2x + 5x-5`

`=(2x^2 - 2x)+(5x-5)`

`= 2x (x-1) + 5 (x-1)`

`=(x-1)(2x+5)`

$\\$

`b,`

`x^2 + 4x - y^2+4`

`=(x^2 + 2 . x . 2+2^2)-y^2`

`=(x+2)^2-y^2`

`=(x+2-y)(x+2+y)`

$\\$

`c,`

`2x^2 - 18`

`= 2 (x^2 - 9)`

`=2 (x^2 - 3^2)`

`=2 (x-3)(x+3)`

$\\$

`d,`

`x^4y^4 +4`

`= [(x^2y^2)^2 + 2 . x^2y^2 . 2 +2^2] - (xy)^2`

`= (x^2y^2 +2)^2 - (xy)^2`

`= (x^2y^2 + 2 - xy) (x^2y^2 +2 +xy)`

$\\$

`e,`

`x^7 +x^2 +1`

`= x^7 + x^6 - x^6 +x^5 - x^5 +x^4 - x^4 +x^3 - x^3 +x^2 + x-x+1`

`= (x^7 + x^6 + x^5) - (x^6 +x^5 +x^4) + (x^4 + x^3 +x^2) - (x^3 +x^2 + x) + (x^2 +x+1)`

`= x^5 (x^2 +  x+1) - x^4 (x^2 +x+1) + x^2 (x^2 +x+1) - x (x^2 +x+1) + 1 (x^2+x+1)`

`= (x^2 +x+1) (x^5 - x^4 +x^2 - x+1)`