\(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\)

\(=x^4+\frac{2}{5}x^2y-\frac{2}{5}x^2y-\left(\frac{2}{5}y\right)^2\)

\(=x^4-\frac{4}{25}y^2\)

a/ \(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)

b/ (x - 3y)(x² + 3xy + 9y²) = x³ - 27y³

c/ (5 + 3x)³ = 125 + 225x + 135x² + 27x³

a) ( x^2 + 2/5y).(x^2-2/5y)

= x^4 - 4/25y^2

b) ( x- 3y)( x^2 + 3xy + 9y^2)

= x^3 - 27y^3

c) (5+3x)^3

= 125 + 225x + 135x^3 + 27x^3

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

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

Áp dùng hằng đằng thức lập phương của một hiệu

\(\Rightarrow x^3-9y^3\)

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

\(\left(x-3y\right)\cdot\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\cdot\left(x^2+x\cdot3y+\left(3y^2\right)\right)\)=\(x^3-\left(3y\right)^3\)

a) \(\left(2x^3-y^2\right)^3\)

\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)

\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)

\(=8x^9-12x^6y^2+6x^3y^4-y^6\)

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

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

\(=x^3-27y^3\)

c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)

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

\(=x^2+4xy+4y^2-z^2\)

d) \(\left(2x^3y-0,5x^2\right)^3\)

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

\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)

e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)

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

\(=4x^4+9x^2-12x^2-27\)

\(=4x^4-3x^2-27\)

f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)

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

\(=8x^3-1\)

\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)

\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)

xin hỏi bạn có viết lộn không, vế trái không có Z mà tại sao vế phải lại xuất hiện Z vậy

a) \(\left(2x^3y-0,5x^2\right)^3\)

\(=\left(2x^3y\right)^3-3\left(2x^3y\right)^20,5x^2+3.2x^3y\left(0,5x^2\right)^2-\left(0,5x^2\right)^3\)

\(=8x^9y^3-6x^8y^2+1,5x^7y-0,125x^6\)

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

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

\(=x^3-27y^3\)

c) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)

\(=x^3-3^3\)

\(=x^3-27.\)

a,\(\left(2x^3y-0,5x^2\right)^3=\left(2x^3y\right)^3-3.\left(2x^3y\right)^2.\left(0,5x^2\right)+3.\left(0,5x^2\right)^2.\left(2x^3y\right)-\left(0,5x^2\right)^3\)

\(=8x^9y^3-6x^8y^2+\frac{3}{2}x^7y-\frac{1}{8}x^6\)

b,\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)

\(=x^3-\left(3y\right)^3=x^3-27y^3\)

\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+3.x^2+3^2\right]\)

\(=\left(x^2\right)^3-3^3=x^6-27\)

a: \(\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)=\left(2x+3y\right)^2:\left(2x+3y\right)=2x+3y\)

d: \(\left(x^2+6xy+9y^2\right):\left(x+3y\right)=\left(x+3y\right)^2:\left(x+3y\right)=x+3y\)

e: \(\dfrac{64y^3-27}{4y-3}=\dfrac{\left(4y-3\right)\left(16y^2+12y+9\right)}{4y-3}=16y^2+12y+9\)

a, \(4x^2+12xy+9y^2=\left(2x+3y\right)^2\)

\(\Rightarrow\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)\)

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

b,\(x^2+6xy+9y^2=\left(x+3y\right)^2\)

\(\Rightarrow\left(x^2+6xy+9y^2\right):\left(x+3y\right)\\ =\left(x+3y\right)^2:\left(x+3y\right)\\ =x+3y\)

c, \(64y^3-27=\left(4y-3\right)\left(16y^2+12y+9\right)\)

\(\Rightarrow\left(64x^3-27\right):\left(4y-3\right)\\ =\left[\left(4y-3\right)\left(16x^2+12x+9\right)\right]:\left(4y-3\right)\\ =16x^2+12x+9\)

=x³+4³

=x³+64