\(C=\left(\frac{x}{2}-y\right)^3-3.2.\left(\frac{x}{2}-y\right)^2+3.2^2\left(\frac{x}{2}-y\right)-2^3=\left(\frac{x}{2}-y-2\right)^3\)

với \(x=206,y=1\Rightarrow C=\left(\frac{x}{2}-y-2\right)^3=\left(\frac{206}{2}-1-2\right)^3=1000\text{ }000\)

a, (\(x\) + y).(\(x\) + y)² - 3\(xy\).(\(x\) + y) 

= (\(x+y\))³ - 3\(x^2\)y - 3\(xy^2\)

= \(x^3\) + 3\(x^2\).y + 3\(xy^2\) + y³ - 3\(x^2\).y  - 3\(xy^2\)

= \(x^3\) + y³ 

b, (\(x-y\)).(\(x-y\))² - 3\(xy\).(\(x-y\)) 

=    (\(x\) - y)³ - 3\(x^2\).y + 3\(xy^2\)

= \(x^3\) - 3\(x^2\)y + 3\(xy^2\) - y³ - 3\(x^2\)y + 3\(xy^2\)

= \(x^3\) - 6\(x^2\)y + 6\(xy^2\) - y³

c, (\(x\) - 2y)² + 4y²

=  \(x^2\) - 4\(xy\) + 4y² + 4y²

=  \(x^2\) - 4\(xy\) + 8y²

gúp mình với

A = x ( x + y ) - y ( x + y )

A = ( x + y ) ( x - y )

A = x\(^2\) - y\(^2\)

Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có 

\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)

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

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

\(=x^2-y^2\)

Thay \(x=-\dfrac{1}{2}\) và \(y=-2\) vào biểu thức \(A\), ta có:

\(A=\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2\)

\(=\dfrac{1}{4}-4\)

\(=-\dfrac{15}{4}\)

\(a,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=4xy\\ b,\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(x-y\right)=\left(x+y-x+y\right)^2=4y^2\\ c,\left(x^2-1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x^3+1\right)\\ =x^4-x^3+x-1\)

a. (x + y)² - (x - y)²

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

= 2y . 2x

= 4xy

b. (x + y)² + (x - y)² - 2(x + y)(x - y)

= (x² + 2xy + y²) + (x² - 2xy + y²) - 2(x² - y²)

= x² + 2xy + y² + x² - 2xy + y² - 2x² + 2y²

= x² + x² - 2x² + 2xy - 2xy + y² + y² + 2y²

= 4y²

c. (x² - 1)(x² - x + 1)

= x⁴ - x³ + x² - x² + x - 1

= x⁴ - x³ + x - 1

a: \(\left(x+y\right)^2-\left(x-y\right)^2=4xy\)

b: \(\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2=\left(x+y-x+y\right)^2=4y^2\)

c: \(\left(x^2-1\right)\cdot\left(x^2-x+1\right)\)

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

\(=x^4-x^3+x-1\)

\(\left(a\right):\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)\\ =x^2+2xy+y^2-x^2+2xy-y^2\\ =4xy\)

\(\left(b\right):\left(x-y-z\right)^2+\left(x+y+z\right)^2\\ =\left[\left(x-y\right)-z\right]^2+\left[\left(x+y\right)+z\right]^2\\ =\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x+y\right)^2+2z\left(x+y\right)+z^2\\ =x^2-2xy+y^2-2xz+2yz+z^2+x^2+2xy+y^2+2xz+2yz+z^2\\ =2x^2+2y^2+2z^2+4yz\)

\(\left(c\right):\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =\left(2y\right)^2=4y^2\)

\(=\dfrac{2}{x+y}\cdot\dfrac{\sqrt{3}\left(x+y\right)}{2}=\sqrt{3}\)

A=\(\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(1+x\right)}\)\(-\frac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)

A=\(\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(1+x\right)\left(1-y\right)\left(x+y\right)}\)

A=\(\frac{x^2+x^3-y^2+y^3-x^2y^2\left(x+y\right)}{\left(1+x\right)\left(1-y\right)\left(x+y\right)}\)

A=\(\frac{\left(x+y\right)\left(x-y\right)+\left(x+y\right)\left(x^2-xy+y^2\right)-x^2y^2\left(x+y\right)}{\left(1+x\right)\left(1+y\right)\left(x+y\right)}\)

A=\(\frac{\left(x+y\right)\left(x-y+x^2-xy+y^2-x^2y^2\right)}{\left(x+y\right)\left(x+1\right)\left(1-y\right)}\)

A=\(\frac{x\left(x+1\right)-y\left(x+1\right)+y^2\left(1-x\right)\left(1+x\right)}{\left(x+1\right)\left(1-y\right)}\)

A=\(\frac{\left(x+1\right)\left(x-y+y^2-y^2x\right)}{\left(x+1\right)\left(1-y\right)}\)

A=\(\frac{-y\left(1-y\right)+x\left(1-y\right)\left(1+y\right)}{\left(1-y\right)}\)

A=\(\frac{\left(1-y\right)\left(-y+x+xy\right)}{1-y}\)=\(x-y+xy\)

D

D