\(=\left(-5a\right)^3+3.\left(-5a\right)^2+3.\left(-5a\right)+1\)

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

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: \(64-96a+48a^2-8a^3\)

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

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

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

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

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

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

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

4x²y-8xy²+18x²y²=2xy(2x-4y+9xy)

\(4x^2y-8xy^2+18x^2y^2=2xy\left(2x-4y+9xy\right)\)

a: Sửa đề: \(-9x^2+12xy-4y^2\)

\(=-\left(9x^2-12xy+4y^2\right)\)

\(=-\left\lbrack\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2\right\rbrack=-\left(3x-2y\right)^2\)

b: \(-125a^3+75a^2-15a+1\)

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

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

c: \(64-96a+48a^2-8a^3\)

\(=8\left(8-12a+6a^2-a^3\right)=8\left(2-a\right)^3\)

d: \(-\frac18m^3n^6-\frac{1}{27}=-\left\lbrack\left(\frac12mn^2\right)^3+\left(\frac13\right)^3\right\rbrack\)

\(=-\left(\frac12mn^2+\frac13\right)\left(\frac14m^2n^4-\frac12mn^2\cdot\frac13+\frac19\right)\)

\(=-\left(\frac12mn^2+\frac13\right)\left(\frac14m^2n^4-\frac16mn^2+\frac19\right)\)

\(1,=8xy+14y^2-4xz-7yz\\ 2,=y\left(4x^2-12x+9\right)=y\left(2x-3\right)^2\\ 3,\Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Câu 1: \(\left(2y-z\right)\left(4x+7y\right)=8xy-4xz+14y^2-7yz\)

câu 2: \(4x^2y-12xy+9y=y\left(4x^2-12x+9\right)\)

câu 3: \(\left(x-2\right)\left(x+3\right)+x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\\ \Leftrightarrow2\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

a) 20x - 5y

= 5(4x - y)

b) 5x(x - 1)- 3x(x - 1)

= 2x(x - 1)

c) x(x + y) - 6x - 6y

= x(x + y) - (6x + 6y)

= x(x + y) - 6(x + y)

= (x + y)(x - 6)

d) 6x³ - 9x²

= 3x²(2x - 3)

e) 4x²y - 8xy² + 10x²y²

= 2xy(2x - 4y + 5xy)

g) 20x²y - 12x³

= 4x²(5y - 3x)

h) 8x⁴ + 12x²y⁴ - 16x³y⁴

= 4x²(2x² + 3y⁴ - 4xy⁴)

k) 4xy² + 8xyz

= 4xy(y + 2z)

giúp mình vs ạ c ơn nhiều

\(=6x\left(2y-1\right)-\left(2y+1\right)\)

bn xem lại có sai đề k nha, mk thấy hơi kì

=3y*x^2-3y*3yz+3y*4x

=3y*(x^2-3yz+4x)

Ta có

x 3 y 3   +   6 x 2 y 2   +   12 x y   +   8 =   ( x y ) 3   +   3 ( x y ) 2 . 2   +   3 x y . 2 2   +   2 3   =   ( x y   +   2 ) 3

Đáp án cần chọn là: A

TA có: \(3x^2-12xy+12y^2-27\)

\(=3\left(x^2-4xy+4y^2-9\right)\)

\(=3\left\lbrack\left(x-2y\right)^2-9^{}\right\rbrack\)

=3(x-2y-3)(x-2y+3)