giúp em với ạ TvT em đang cần gấp lắm huhu

b)

Sửa đề: \(125a^3+75a^2+15a+1\)

Ta có: \(125a^3+75a^2+15a+1\)

\(=\left(5a\right)^3+3\cdot\left(5a\right)^2\cdot1+3\cdot5a\cdot1^2+1^3\)

\(=\left(5a+1\right)^3\)

c) Ta có: \(64-96a+48a^2-8a^3\)

\(=-\left(8a^3-48a^2+96a-64\right)\)

\(=-\left[\left(8a^3-64\right)-48a\left(a-2\right)\right]\)

\(=-\left[\left(2a-4\right)\left(4a^2+8a+16\right)-48a\left(a-2\right)\right]\)

\(=-\left[\left(a-2\right)\left(8a^2+16a+32-48a\right)\right]\)

\(=-\left(a-2\right)\left(8a^2-32a+32\right)\)

\(=-8\left(a-2\right)\left(a^2-4a+4\right)\)

\(=-8\left(a-2\right)^3\)

64-96a+48a²-8a^3= 4³ - 3.4².21 +3.4.(2a)² - (2a)³= (4-2a)³

a) (4- 2a)³

\(16^4+y^4=\left[\left(y^2\right)^2+2.y^2.16^2+\left(16^2\right)^2\right]-2.y^2.16^2=\left(y^2+16^2\right)^2-2.y^2.16^2\)

b tự tính tiếp nhé

ý b tương tự. ( gợi ý: thêm bớt hạng tử 16y^4 )

\(y^8+64\)

\(=\left(y^4\right)^2+2\cdot y^4\cdot8+8^2-2\cdot y^4\cdot8\)

\(=\left(y^4+8\right)^2-16y^4\)

\(=\left(y^4+8\right)^2-\left(4y^2\right)^2\)

\(=\left(y^4+8-4y^2\right)\left(y^4+8+4y^2\right)\)

a kudo shinichi làm rồi đó

\(x^2-4y^2+16x+64=\left(x^2+16x+64\right)-4y^2\)

\(=\left(x+8\right)^2-\left(2y\right)^2\)

\(=\left(x-2y+8\right)\left(x+2y+8\right)\)

\(64-x^2-y^2+2xy=64-\left(x^2-2xy+y^2\right)\)

\(=8^2-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\)

−(x−y−8)(x−y+8)

Đa thức không phân tích được bạn ơi

1) \(64-y^2=8^2-y^2=\left(8-y\right)\left(8+y\right)\)

2) \(81-x^2=9^2-x^2=\left(9-x\right)\left(9+x\right)\)

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

4) \(144-b^2=12^2-b^2=\left(12-b\right)\left(12+b\right)\)