13! = 1. 2. 3. 4. 5. 6. 7. 8. 9. 11. 12. 13 = 6 227 020 800

Số 13! có tận cùng 

là hai 

chữ số 0

mik chịu sẵn nha

Hỏa Long

12.

\(hpt\Leftrightarrow\left\{{}\begin{matrix}x^2+7=4y^2+4y\\\left(x+y\right)\left(x+2y+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+7=4y^2+4y\\\left[{}\begin{matrix}x=-y\\x=-2x-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-y\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-2x-1\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2+7=4y^2+4y\\x=-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(y-3\right)\left(y+7\right)=0\\x=-y\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=3\\x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y=-7\\x=7\end{matrix}\right.\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-2x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y^2+4y+1+7=4y^2+4y\\x=-2x-1\end{matrix}\right.\text{ vô nghiệm}\)

Vậy hệ pt đã cho có 2 nghiệm ...

13.

\(\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3+\left(x-5\right)^2+\left(y+5\right)^2=55\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3+x^2+y^2-10x+10y+50=55\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3=10x-10y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3=2.5\left(x-y\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3=2.\left(x^2+y^2\right)\left(x-y\right)\left(1\right)\end{matrix}\right.\)

Từ \(\left(1\right)\) ta có: \(x^3-2x^2y+2xy^2-4y^3=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x^2+2y^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2y=0\\x^2+2y^2=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x^2+y^2=5\\x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y^2=5\\x=2y\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x=-2\end{matrix}\right.\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x^2+y^2=5\\x^2+2y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\y^2=-5\end{matrix}\right.\text{ vô nghiệm}\)

Vậy hệ pt đã cho có 2 nghiệm \(\left(x;y\right)\in\left\{\left(1;2\right);\left(-1;-2\right)\right\}\)

10.

\(\left\{{}\begin{matrix}2x^2-3xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2xy-xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x-y+1\right)=0\\x^2+x+1=y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\y=2x+1\end{matrix}\right.\\x^2+x+1=y^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=y^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=y^2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=x^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=\left(2x+1\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\3x\left(x+1\right)=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=1\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2x+1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x=-1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Số mới là: 92 - 28 = 64

Số ban đầu là; 64 - 18 = 46

9x9=81