\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

Bài giải:

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

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

\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\\ =3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\\=-10xy+2y^2 \)

b) \(3\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(=3\left(4x^2+20x+25\right)-3\left(1+4x\right)\left(1-4x\right)\)

\(=12x^2+60x+75-3\left(1-16x^2\right)\)

\(=12x^2+60x+75-3+48x^2\)

\(=60x^2+60x+72\)

câu b là \(^{2\left(2x+5\right)^2-\left(4x+1\right)\cdot\left(1-4x\right)}\) xin lỗi mk chép đề sai ạ

Bài làm

a) 2(x + y)³ + 2(x - y)³ 

= 2[(x + y)³ + (x - y)³]

= 2[x³ + 3x²y + 3xy² + y³ + x³ - 3x²y + 3xy² - y³]

= 2[(x³ + x³) + (3x²y - 3x²y) + (3xy² + 3xy²) + (y³ - y³)]

= 2[2x³ + 6xy²]

= 4x³ + 12xy²

b)uhm... Mình sửa đề chút, thay vì là -3(x + y)²(x - y) thì mình sẽ thành +3(x + y)²(x - y)

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

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

= -[(x + y) - (x - y)]³ 

= -[x + y - x + y ]³

= -[y]³ 

= -y

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

`=(y(y-2x))/3`

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

`=(-(x^2-3x+x-3))/((x-1)(x+1))`

`=(-x(x-3)+x-3)/((x-1)(x+1))`

`=((x-3)(1-x))/((x-1)(x+1))`

`=(3-x)/(1+x)`

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

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

b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)

\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)

\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)

\(=-x^2+18xy\)

c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)

\(=\left(2a-3b\right)^2-16c^2\)

\(=4a^2-12ab+9b^2-16c^2\)

a: \(A=\dfrac{x^5}{x^3}\cdot\dfrac{y^{-2}}{y}=x^2\cdot y^{-1}=\dfrac{x^2}{y}\)

b: \(B=\dfrac{x^2\cdot y^{-3}}{x^3\cdot y^{-12}}=\dfrac{x^2}{x^3}\cdot\dfrac{y^{-3}}{y^{-12}}=\dfrac{1}{x}\cdot y^{-3+12}=\dfrac{y^9}{x}\)

a) \(A=\dfrac{x^5y^{-2}}{x^3y}=\dfrac{x^5}{x^3}.\dfrac{1}{y^{2-1}}=x^{5-3}y^{-1}=x^2y^{-1}\).

b) \(B=\dfrac{x^2y^{-3}}{\left(x^{-1}y^4\right)^{-3}}=\dfrac{x^2y^{-3}}{x^3y^{-12}}=x^{2-3}y^{-3-\left(-12\right)}=\dfrac{1}{xy^9}\)

a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

\(=2x^2-4xy+\dfrac{15}{4}y^2\)

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

\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(=2x^2+2x+13-2x^2+2\)

=2x+15

a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)

b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(=2x+15\)

a; \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

= \(x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

= \(2x^2-4xy+\dfrac{15}{4}y^2\)

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

= \(x^2-4x+4+x^2+6x+9-2x^2+2\)

= \(2x+15\)