\(A=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3.5.4=65\)

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

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

\(=4\left(x^2+2xy+y^2-3xy\right)=4\left[\left(x+y\right)^2-3.3\right]=4\left(16-9\right)=28\)

Lời giải:

Theo hằng đẳng thức đáng nhớ:

$x^3+y^3=(x+y)^3-3xy(x+y)=4^3-3.3.4=28$

 a) Ta thấy \(xy=\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\dfrac{3^2-5}{2}=2\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\) \(=3\left(5-2\right)=9\)

 b) Ta thấy \(xy=\dfrac{-\left(x-y\right)^2+\left(x^2+y^2\right)}{2}=\dfrac{15-5^2}{2}=-5\)

\(\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\) \(=5\left(15-5\right)=50\)

Bài 5

a) A = -x³ + 6x² - 12x + 8

= -x³ + 3.(-x)².2 - 3.x.2² + 2³

= (-x + 2)³

= (2 - x)³

Thay x = -28 vào A ta được:

A = [2 - (-28)]³

= 30³

= 27000

b) B = 8x³ + 12x² + 6x + 1

= (2x)³ + 3.(2x)².1 + 3.2x.1² + 1³

= (2x + 1)³

Thay x = 1/2 vào B ta được:

B = (2.1/2 + 1)³

= 2³

= 8

Bài 6

a) 11³ - 1 = 11³ - 1³

= (11 - 1)(11² + 11.1 + 1²)

= 10.(121 + 11 + 1)

= 10.133

= 1330

b) Đặt B =  x³ - y³ = (x - y)(x² + xy + y²)

= (x - y)(x² - 2xy + y² + 3xy)

= (x - y)[(x - y)² + 3xy]

Thay x - y = 6 và xy = 9 vào B ta được:

B = 6.(6² + 3.9)

= 6.(36 + 27)

= 6.63

= 378

Bài 6 :

a) \(11^3-1=\left(11-1\right)\left(11^2+11+1^2\right)\)

\(\)\(=10.\left(121+12\right)\)

\(=10.133\)

\(=1330\)

b) \(\left\{{}\begin{matrix}x-y=6\\xy=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-2xy=36\\xy=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-2.18=36\\xy=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=72\\xy=9\end{matrix}\right.\)

Ta có :

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

\(=6.\left(72+9\right)\)

\(=6.81\)

\(=486\)

\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)

\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)

\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)

a) \(11^3-1\)

\(=11^3-1^3\)

\(=\left(11-1\right)\left(11^2+11\cdot1+1^2\right)\)

\(=10\cdot\left(121+11+1\right)\)

\(=10\cdot\left(132+1\right)\)

\(=10\cdot133\)

\(=1330\)

b) Ta có:
\(x^3-y^3\)

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

Thay \(x-y=6\) và \(xy=20\) ta có:

\(6^3+3\cdot20\cdot6=216+60\cdot6=216+360=576\)

a: 11^3-1=(11-1)(11^2+11+1)

=10*(121+12)

=10*133=1330

b: x^3-y^3=(x-y)^3+3xy(x-y)

=6^3+3*20*6

=216+360

=576

`#3107.101107`

`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`

Ta có:

`x - y - 1 = 0`

`=> x - y = 1`

`D = x^3 - y^3 - 3xy`

`= (x - y)(x^2 + xy + y^2) - 3xy`

`= 1 * (x^2 + xy + y^2) - 3xy`

`= x^2+ xy + y^2 - 3xy`

`= x^2 - 2xy + y^2`

`= x^2 - 2*x*y + y^2`

`= (x - y)^2`

`= 1^2 = 1`

Vậy, với `x - y = 1` thì `D = 1`

________

`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`

`x + y = 5`

`=> (x + y)^2 = 25`

`=> x^2 + 2xy + y^2 = 25`

`=> 2xy = 25 - (x^2 + y^2)`

`=> 2xy = 25 - 17`

`=> 2xy = 8`

`=> xy = 4`

Ta có:

`E = x^3 + y^3`

`= (x + y)(x^2 - xy + y^2)`

`= 5 * [ (x^2 + y^2) - xy]`

`= 5 * (17 - 4)`

`= 5 * 13`

`= 65`

Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`

________

`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`

Ta có:

`x - y = 4`

`=> (x - y)^2 = 16`

`=> x^2 - 2xy + y^2 = 16`

`=> (x^2 + y^2) - 2xy = 16`

`=> 2xy = (x^2 + y^2) - 16`

`=> 2xy = 26 - 16`

`=> 2xy = 10`

`=> xy = 5`

Ta có:

`F = x^3 - y^3`

`= (x - y)(x^2 + xy + y^2)`

`= 4 * [ (x^2 + y^2) + xy]`

`= 4 * (26 + 5)`

`= 4*31`

`= 124`

Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`

\(\text{a) Ta có:}xy=1\Rightarrow\hept{\begin{cases}2xy=2\\-2xy=-2\end{cases}}\)

\(\text{Ta lại có: }x^2+y^2=2\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=2+2=4\\x^2+y^2-2xy=2-2=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=4\\\left(x-y\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=\pm2\\x-y=0\end{cases}}}\)

\(\text{b) Ta có: }x+y=5\)

\(\Rightarrow\left(x+y\right)^2=25\)

\(\Rightarrow x^2+2xy+y^2=25\)

\(\Rightarrow x^2+4+y^2=25\)

\(\Rightarrow x^2+y^2=21\)

\(\text{b) Ta có: }x^2+y^2=21\)

\(\Rightarrow x^2-2xy+y^2=21-2xy\)

\(\Rightarrow\left(x-y\right)^2=21-4\)

\(\Rightarrow\left(x-y\right)^2=17\)

\(\Rightarrow x-y=\pm\sqrt{17}\)