Đáp án:
a.$ A=(a^2+1)(b^2+1)$
b.$B=(1-y)(x-1)(x+xy+y)$
c.$C=(x-2y-2)(x+2y)$
d.$D=(a-b)^3(a+b)$
Giải thích các bước giải:
a.Ta có:
$A=(ab-1)^2+(a+b)^2$
$\to A=(ab)^2-2ab+1+a^2+2ab+b^2$
$\to A=(ab)^2+a^2+b^2+1$
$\to A=a^2(b^2+1)+(b^2+1)$
$\to A=(a^2+1)(b^2+1)$
b.Ta có:
$B=x^2+y^2-x^2y^2+xy-x-y$
$\to B=(x^2-x^2y^2)+(y^2+xy)-(x+y)$
$\to B=x^2(1-y^2)+y(x+y)-(x+y)$
$\to B=x^2(1-y)(1+y)+(x+y)(y-1)$
$\to B=x^2(1-y)(1+y)-(x+y)(1-y)$
$\to B=(1-y)(x^2(1+y)-(x+y))$
$\to B=(1-y)(x^2+x^2y-x-y)$
$\to B=(1-y)((x^2-x)+(x^2y-y))$
$\to B=(1-y)(x(x-1)+y(x^2-1))$
$\to B=(1-y)(x(x-1)+y(x-1)(x+1))$
$\to B=(1-y)(x-1)(x+y(x+1))$
$\to B=(1-y)(x-1)(x+xy+y)$
c.Ta có:
$C=x^2-2x-4y^2-4y$
$\to C=(x^2-2x+1)-(4y^2+4y+1)$
$\to C=(x-1)^2-(2y+1)^2$
$\to C=(x-1-2y-1)(x-1+2y+1)$
$\to C=(x-2y-2)(x+2y)$
d.Ta có:
$D=a(a+2b)^3-b(2a+b)^3$
$\to D=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)$
$\to D=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4$
$\to D=a^4-2a^3b+2ab^3-b^4$
$\to D=(a^4-b^4)-(2a^3b-2ab^3)$
$\to D=(a^2+b^2)(a-b)(a+b)-2ab(a^2-b^2)$
$\to D=(a^2+b^2)(a-b)(a+b)-2ab(a-b)(a+b)$
$\to D=(a^2+b^2-2ab)(a-b)(a+b)$
$\to D=(a-b)^2(a-b)(a+b)$
$\to D=(a-b)^3(a+b)$
