Đáp án:
Giải thích các bước giải:
1,ta có:
(x+1)(x+2)..........(x+23)(x+24)=-420
⇒[(x+1)(x+24)][(x+2)(x+23)].........................[(x+12)(x+13)]=-420
mà
(x+1)(x+24)=$x^{2}$ +25x+25
(x+2)(x+23)=$x^{2}$ +25x+25
.....................
(x+12)(x+13)=$x^{2}$ +25x+25
có 12 cặp như thế
=>$(x^2+25x+25)^{12}=-420
mà $(x^2+25x+25)^{12}≥0
=>$(x^2+25x+25)^{12}=-420(VL)
=>x∈Ф
2,
ta có:
xy+3x-y-3=3
⇔x(y+3)-(y+3)=3
⇔(y+3)(x-1)=3
⇔\(\left[ \begin{array}{l}x-1=3\\y+3=1\end{array} \right.\)
\(\left[ \begin{array}{l}x-1=-3\\y+3=-1\end{array} \right.\)
\(\left[ \begin{array}{l}x-1=1\\y+3=3\end{array} \right.\)
\(\left[ \begin{array}{l}x-1=-1\\y+3=-3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\y=-2\end{array} \right.\)
\(\left[ \begin{array}{l}x=-2\\y=-4\end{array} \right.\)
\(\left[ \begin{array}{l}x=2\\y=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=0\\y=-6\end{array} \right.\)
