+2y là ở trong ngoặc hay ngoài ngoặc á bạn?

Nếu để ngoài ngoặc thì nhìn đc ra HĐT hợp lí hơn, nhưng thoi kệ=))

\(\left(y-5\right)\left(25+5y+y^2+2y\right)=\left(y-5\right)\left(y^2+7y+25\right)\)

\(=y^3+7y^2+25y-5y^2-35y-125=y^3+2y^2-10y-125\)

a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)

b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)

a: =x^3+8-1+27x^3=28x^3+7

b: Sửa đề: (2+y)(y^2-2y+4)+(5-y)(25+5y+y^2)

=8+y^3+125-y^3

=133

\(\left(-7+\left|13\right|\right)-\left(13-\left|-7\right|-25\right)-\left(25+\left|10\right|-9\right)\\ =\left(-7+13\right)-\left(13-7-25\right)-\left(25+10-9\right)\\ =\left(-7\right)+13-13+7+25-25-10+9\\ =\left[\left(-7\right)+7\right]+\left(13-13\right)+\left(25-25\right)+\left(7+9-10\right)\\ =0+0+0+\left(16-10\right)\\ =0+6\\ =6\)

a) \(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)

b) \(=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)

c) \(=3x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(3x+5\right)\)

d) \(=\left(x+y\right)^2-25=\left(x+y-5\right)\left(x+y+5\right)\)

e) \(=x\left(x^2-11x+30\right)\)

f) \(=x\left(x-3\right)+6\left(x-3\right)=\left(x-3\right)\left(x+6\right)\)

\(-3xy^2+x^2y^2-5x^2y\)

\(=-xy\left(3y+xy-5x\right)\)

\(x\left(y-1\right)+3\left(y^3+2y+1\right)\)

\(=3y^3+6y+3+xy-x\)

Xem lại nhé ko phân tích được

\(12xy^2-12xy+3x\)

\(=3x\left(4y^2-4y+1\right)\)

\(=3x\left(2y-1\right)^2\)

\(10x^2\left(x+y\right)-5\left(2x+2y\right)y^2\)

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

\(=10\left(x+y\right)\left(x-y\right)\left(x+y\right)\)

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

a) x=3 ; y=8
b) x=4 ; y=0
c) x=3 ; y=0
d) x=3 ; y=0