Question

[Trần Khánh Huyền]

mn ơi giúp hộ mình vs a

Answer 1

[missing you =]

a,\(5x+10y=5\left(x+2y\right)\)

b,\(=3xy\left(x+3yz\right)\)

c,\(=\left(x+1\right)\left(5-3x-3\right)=\left(x+1\right)\left(2-3x\right)\)

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

e,\(=y\left(xy+3x^2+y^2\right)\)

f,\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

g,\(=x^2-x+\dfrac{1}{4}-\left(\dfrac{5}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=\left(x-3\right)\left(x+2\right)\)

h,\(=x^2+9x+\dfrac{81}{4}-\left(\dfrac{7}{2}\right)^2=\left(x+\dfrac{9}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(x+1\right)\left(x+8\right)\)

i,\(=x^2-10x+25-4^2=\left(x-5\right)^2-4^2=\left(x-1\right)\left(x-9\right)\)

k,\(=x^2+x+\dfrac{1}{4}-\left(\dfrac{7}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(x-3\right)\left(x+4\right)\)

l.\(=3\left(x^2+\dfrac{8}{3}x+\dfrac{4}{3}\right)=3\left(x^2+\dfrac{8}{3}x+\dfrac{16}{9}-\dfrac{4}{9}\right)\)

\(=3[\left(x+\dfrac{4}{3}\right)^2-\left(\dfrac{2}{3}\right)^2]=3\left(x+\dfrac{2}{3}\right)\left(x+2\right)\)

Answer 2

[Út Thảo]

A, 5(x+2y)

B, 3xy(x+3yz)

C, (x+1)[5-3(x+1)]

D, (x-y)(2+3x)

E, y(xy+3x^2+y^2)

F ,(x+y)(x-y-1)

G, (x-3)(x+2)

H, (x+1)(x+8)

I, (x-9)(x-1)

K, (x-3)(x+4)

L, (x+2)(3x+2)