Bài 2b:
\(n^2+2n-11⋮n+2\)
\(\Leftrightarrow n\left(n+2\right)-11⋮n+2\)
\(\Leftrightarrow-11⋮n+2\)
\(\Rightarrow n+2\in\text{Ư}\left(11\right)=\left\{\pm1;\pm11\right\}\)
Ta có bảng:
| \(n+2\) | \(1\) | \(-1\) | \(11\) | \(-11\) |
| \(n\) | \(-1\) | \(-3\) | \(9\) | \(-13\) |
Vậy \(n\in\left\{-1;-3;9;-13\right\}\)
\(\#PeaGea\)
Bài 4:
a:
Vì \(\dfrac{S_{AOD}}{S_{DOC}}=\dfrac{5}{8}\) nên \(\dfrac{OA}{OC}=\dfrac{5}{8}\)
=>\(S_{ABO}=\dfrac{5}{8}\cdot S_{BOC}\)
=>\(S_{ABO}=\dfrac{5}{8}\cdot10=6,25\left(cm^2\right)\)
b:ΔABC vuông tại A
=>\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=75\left(cm^2\right)\)
Vì M là trung điểm của BC
nên \(S_{AMC}=\dfrac{1}{2}\cdot75=37,5\left(cm^2\right)\)
Vì \(CD=\dfrac{1}{3}CA\)
nên \(S_{MDC}=\dfrac{1}{3}\cdot S_{MAC}=\dfrac{1}{3}\cdot37,5=12,5\left(cm^2\right)\)
Bài 2:
a: x+6=y(x-1)
=>x-1+7=y(x-1)
=>x-1-y(x-1)=-7
=>(x-1)(1-y)=-7
=>(x-1)(y-1)=7
=>\(\left(x-1\right)\left(y-1\right)=1\cdot7=\left(-1\right)\cdot\left(-7\right)=\left(-7\right)\cdot\left(-1\right)=7\cdot1\)
=>\(\left(x-1;y-1\right)\in\left\{\left(1;7\right);\left(-1;-7\right);\left(-7;-1\right);\left(7;1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(2;8\right);\left(0;-6\right);\left(-6;0\right);\left(8;2\right)\right\}\)
mà x,y là các số tự nhiên
nên \(\left(x;y\right)\in\left\{\left(2;8\right);\left(8;2\right)\right\}\)
