giải hệ phương trình
1, \(\left\{{}\begin{matrix}2x^2+3y=17\\3x^2-2y=6\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-1\right|=2\\4\left|x-1\right|+3\left|y-1\right|=7\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=2\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}x+y=2\\\left|2x-3y\right|=1\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}2x-y=1\\\left|x-y\right|=\left|2y-1\right|\end{matrix}\right.\)
6,\(\left\{{}\begin{matrix}\left(x-3\right)\left(y+6\right)=xy\\\left(x+2\right)\left(y-2\right)=xy\end{matrix}\right.\)
7 , \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}4x^2-5\left(y+1\right)=\left(2x-3\right)^2\\3\left(7x+2\right)=5\left(2y-1\right)-3x\end{matrix}\right.\)
giải hpt:
a) \(\left\{{}\begin{matrix}4x+9y=6\\3x^2+6xy-x+3y=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(x+y+2\right)\left(2x+2y-1\right)=0\\3x^2-32y^2+5=0\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}2x^2-xy+3y^2=7x+12y-1\\x-y+1=0\end{matrix}\right.\)
Giải hệ pt
a) \(\left\{{}\begin{matrix}x^2+2xy^2=3\\y^3+y+x\left(2xy-1\right)=3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+x^3y-xy^2+xy-y=1\\x^4+y^2-xy\left(2x-1\right)=1\end{matrix}\right.\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\left(x+y\right)^2+3y^2=7\\x+2y\left(x+1\right)=5\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x\left(y-1\right)+2y=x\left(x+1\right)\\\sqrt{2x-1}+xy-3y+1=0\end{matrix}\right.\)
Giải 1 trong 2 bài cũng được. Thanks!!!
Giải hệ
a) \(\left\{{}\begin{matrix}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(6x+5\right)\sqrt{2x+1}-2y-3y^3=0\\y+\sqrt{x}=\sqrt{2x^2+4x-23}\end{matrix}\right.\)
Giải bất pt
\(\dfrac{9}{\left|x-5\right|-3}\ge\left|x-2\right|\)
Giải hệ phương trình:
a) \(\left\{\begin{matrix}2x^2-15xy+4y^2-12x+45y-24=0y^2\\x^2+xy-2y^2-3x-3y=0\end{matrix}\right.\)
b) \(\left\{\begin{matrix}3\left|x-3\right|+5y+9=0\\2x-\left|y+4\right|-7=0\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}2\left|x-6\right|+3\left|y-1\right|=5\\5\left|x-6\right|-4\left|y+1\right|=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2\left|x+y\right|-\left|x-y\right|=9\\3\left|x+y\right|+2\left|x-y\right|+17\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}4\left|x+y\right|+3\left|x-y\right|=8\\3\left|x+y\right|-5\left|x-y\right|=6\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}x^2-xy=24\\2x-3y=1\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}3x-4y+1=0\\xy=3\left(x+y\right)-9\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}2x+3y=5\\3x^2-y^2+2y=4\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?