Question

q.x^2-3x+3y-y^2
r.-a^2+5a-ab+5b
s.x^2-16a^2+y^2-2xy
x.a^2-c^2+(a+b)(c-a)
z.100-(2x-3y)^2

Answer 1

q

\(x^2-3x+3y-y^2\\ =\left(x-y\right)^2-\left(3x-3y\right)\\ =\left(x-y\right)\left(x-y\right)-3\left(x-y\right)\\ =\left(x-y\right)\left(x-y-3\right)\)

r

\(-a^2+5a-ab+5b\\ =-\left(a^2-5a+ab-5b\right)\\ =-\left[a\left(a+b\right)-5\left(a+b\right)\right]\\ =-\left(a-5\right)\left(a+b\right)\)

s

\(x^2-16a^2+y^2-2xy\\ =\left(x^2-2xy+y^2\right)-\left(4a\right)^2\\ =\left(x-y\right)^2-\left(4a\right)^2\\ =\left(x-y-4a\right)\left(x-y+4a\right)\)

 

Answer 2

x

\(a^2-c^2+\left(a+b\right)\left(c-a\right)\\ =\left(a-c\right)\left(a+c\right)+\left(a+b\right)\left(c-a\right)\\ =\left(a-c\right)\left(a+c\right)-\left(a+b\right)\left(a-c\right)\\ =\left(a-c\right)\left(a+c-a-b\right)\\ =\left(a-c\right)\left(c-b\right)\)

z

\(100-\left(2x-3y\right)^2\\ =10^2-\left(2x-3y\right)^2\\ =\left(10-2x+3y\right)\left(10+2x-3y\right)\)

Answer 3

Lời giải:

q. $=(x^2-y^2)-(3x-3y)=(x-y)(x+y)-3(x-y)=(x-y)(x+y-3)$

r. $=-(a^2+ab)+(5a+5b)=-a(a+b)+5(a+b)=(a+b)(-a+5)$

s. $=(x^2-2xy+y^2)-16a^2=(x-y)^2-(4a)^2=(x-y-4a)(x-y+4a)$

x. $=(a-c)(a+c)-(a+b)(a-c)=(a-c)[(a+c)-(a+b)]=(a-c)(c-b)$

z. $=10^2-(2x-3y)^2=(10-2x+3y)(10+2x-3y)$