a, x + y = 3 => (x + y)² = 9 <=> x² + 2xy + y² = 9 <=>  5 + 2xy = 9 <=> 2xy = 4 <=> xy = 2

Ta có: x³ + y³ = (x + y)(x² - xy + y²) = 3 . (5 - 2) = 3 . 3 = 9

b, x - y = 5 => (x - y)² = 25 <=> x² - 2xy + y² = 25 <=> 15 - 2xy = 25 <=> -2xy = 10 <=> xy = -5

Ta có: x³ - y³ = (x - y)(x² + xy + y²) = 5 . (15 - 5) = 5 . 10 = 50 

a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)

\(x^5+y^5=\left(x+y\right)\left[x^4-x^3y+x^2y^2-xy^3+y^4\right]\)

\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)

b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)

\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)​​

​\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)​​

\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)

toi can giup

a)  \(x+y=3\)

\(\Rightarrow\)\(\left(x+y\right)^2=9\)

\(\Leftrightarrow\)\(x^2+y^2+2xy=9\)

\(\Leftrightarrow\)\(2xy=4\)  do x² + y² = 5

\(\Leftrightarrow\)\(xy=2\)

   \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=3^3-3.2.3=9\)

b) bạn làm tương tự

\(a,x+y=3\Rightarrow\left(x+y\right)^2=9\Rightarrow x^2+2xy+y^2=9\Rightarrow2xy=4\Leftrightarrow xy=2\)

Vì \(\left(x+y\right)=3\Rightarrow\left(x+y\right)^3=27\)

\(\Rightarrow x^3+3x^2y+3xy^2+y^3=27\)

\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\)

\(\Rightarrow x^3+y^3+3.2.3=27\)

\(\Rightarrow x^3+y^3=27-18=9\)

\(b,x-y=5\Rightarrow\left(x-y\right)^2=25\Rightarrow x^2-2xy+y^2=25\Rightarrow2xy=-10\Leftrightarrow xy=-5\)

\(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=5.10=50\)

Bài làm:

a) Ta có: \(x+y=3\Leftrightarrow\left(x+y\right)^2=3^2\Leftrightarrow x^2+2xy+y^2=9\)

\(\Leftrightarrow2xy+5=9\)

\(\Leftrightarrow2xy=4\)

\(\Rightarrow xy=2\)

Từ đó ta có: \(x^3+y^3\)

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

\(=3\left(5-2\right)\)

\(=3.3=9\)

b) Ta có: \(x-y=5\Leftrightarrow\left(x-y\right)^2=25\Leftrightarrow x^2-2xy+y^2=25\)

\(\Leftrightarrow15-2xy=25\)

\(\Leftrightarrow2xy=-10\)

\(\Rightarrow xy=-5\)

Từ đó ta có: \(x^3-y^3\)

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

\(=5\left(15+10\right)\)

\(=5.25=125\)

`a, (x-y)^2 = (x+y)^2 - 4xy = 12^2 - 35 . 4 = 144 - 140 = 4`.

`b, (x+y)^2 = (x-y)^2 + 4xy = 8^2 + 20.4 = 64 + 80 = 144`

`c, x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 5^3 - 3 . 6 . 5 = 125 - 90 = 35`

`d, x^3 - y^3 = (x-y)^3 - 3xy(x-y) = 3^3 - 3 .40 . 3 = 27 - 360 = -333`.

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

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