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

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

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

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

C

-6

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

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

\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)=13.19-\left(-6\right)^2.1=211\)

b)  \(x^2+y^2=\left(x-y\right)^2+2xy=1^1+2.6=13\)

    \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+3.6.1=19\)

   \(x^5-y^5=\left(x^2+y^2\right)\left(x^3-y^3\right)+x^2y^2\left(x-y\right)=13.19+6^2.1=283\)

`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`.

d: x+y=5

nên x=5-y

Ta có: xy=6

=>y(5-y)=6

=>y²-5y+6=0

=>(y-2)(y-3)=0

=>y=2 hoặc y=3

=>x=3 hoặc x=2

a: \(\Leftrightarrow\left(x-3;y+4\right)\in\left\{\left(1;-7\right);\left(-1;7\right);\left(-7;1\right);\left(7;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(4;-11\right);\left(2;3\right);\left(-4;-3\right);\left(10;-5\right)\right\}\)

chịu

Xài trò này chắc Oke :))

a)

Mình nghĩ là \(x^5+y^5\)nhó, nếu đề khác thì comment xuống mình nghĩ cách khác :p

\(49=\left(x+y\right)^2=x^2+y^2+2xy=25+2xy\Rightarrow xy=12\)

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

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

\(=25\cdot7\cdot\left(25-12\right)-12^2\cdot7\)

\(=1267\)

b)

\(xy^6+x^6y=xy\left(x^5+y^5\right)=P\left(x^5+y^5\right)\)

Ta tính \(x^5+y^5\) theo S và P

Dễ có:

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

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

\(=\left(S^2-2P\right)\left(S^3-3SP\right)-S^2P\)

\(=S^5-5S^3P+2SP^2-S^2P\)

Chắc không nhầm lẫn gì ở việc tính toán =)))

\(x\cdot y=6\)

\(\Rightarrow\hept{\begin{cases}x=1\\y=6\end{cases}}\)hoặc \(\hept{\begin{cases}x=-1\\y=-6\end{cases}}\)

hoặc \(\hept{\begin{cases}x=6\\y=1\end{cases}}\)hoặc \(\hept{\begin{cases}x=-6\\y=-1\end{cases}}\)

hoặc \(\hept{\begin{cases}x=2\\y=3\end{cases}}\)hoặc \(\hept{\begin{cases}x=-2\\y=-3\end{cases}}\)

hoặc \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)hoặc \(\hept{\begin{cases}x=-3\\y=-2\end{cases}}\)

còn câu nào nx ko ạ :(

\(\left(2-x\right)\left(y+1\right)=5\)

có 4 trường hợp:

\(\hept{\begin{cases}2-x=5\\y+1=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=0\end{cases}}\)

\(\hept{\begin{cases}2-x=-5\\y+1=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}\)

\(\hept{\begin{cases}2-x=1\\y+1=5\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=4\end{cases}}\)

\(\hept{\begin{cases}2-x=-1\\y+1=-5\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=-6\end{cases}}\)