1. Tìm số nguyên x, biết:
a. (x - 2).(x + 15) = 0
b. (x + 15).(x - 12) = 0
c. (x - 7).(x + 19) = 0
d. (x - 11).(x + 5) = 0
2. Tìm giá trị lớn nhất của biểu thức: M = 2012 - (x - 1)2
3. Tìm giá trị nhỏ nhất của biểu thức: N = /x - 3/ + 10
4. Tìm số nguyên n sao cho: (n - 6) \(⋮\) (n - 4)
5. Tìm số nguyên n sao cho: (n - 5) \(⋮\) (n -2)
Bài 1:
a) \(\left(x-2\right)\left(x+15\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=2\\x=-15\end{matrix}\right.\)
Vậy \(x\in\left\{3;-15\right\}\)
Các phần khác làm tương tự
Bài 2:
Ta có: \(-\left(x-1\right)^2\le0\)
\(\Rightarrow M=2012-\left(x-1\right)^2\le2012\)
Vậy \(MIN_M=2012\) khi \(x=1\)
Bài 3:
Ta có: \(\left|x-3\right|\ge0\)
\(\Rightarrow N=\left|x-3\right|+10\ge10\)
Vậy \(MAX_M=10\) khi \(x=3\)
Bài 4:
Ta có: \(n-6⋮n-4\)
\(\Rightarrow\left(n-4\right)-2⋮n-4\)
\(\Rightarrow2⋮n-4\)
\(\Rightarrow n-4\in\left\{1;-1;2;-2\right\}\)
\(\left[\begin{matrix}n-4=1\\n-4=-1\\n-4=2\\n-4=-2\end{matrix}\right.\Rightarrow\left[\begin{matrix}n=5\\n=3\\n=6\\n=2\end{matrix}\right.\)
Vậy \(n\in\left\{5;3;6;2\right\}\)
Bài 5: Tương tự bài 4
Bài 1:
b)\(\left(x+15\right)\left(x-12\right)=0\)
\(\Rightarrow\left[\begin{matrix}x+15=0\\x-12=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=-15\\x=12\end{matrix}\right.\)
c)\(\left(x-7\right)\left(x+19\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-7=0\\x+19=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=7\\x=-19\end{matrix}\right.\)
d)\(\left(x-11\right)\left(x+5\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-11=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
Bài 5:
\(\frac{n-5}{n-2}=\frac{n-2-3}{n-2}=\frac{n-2}{n-2}-\frac{3}{n-2}=1-\frac{3}{n-2}\in Z\)
\(\Rightarrow3⋮n-2\Rightarrow n-2\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow n\in\left\{3;1;5;-1\right\}\)
1.
a. (x - 2)(x + 15) = 0
\(\Rightarrow\left\{\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\x=-15\end{matrix}\right.\)
b. (x + 15)(x - 12) = 0
\(\Rightarrow\left\{\begin{matrix}x+15=0\\x-12=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-15\\x=12\end{matrix}\right.\)
c. (x - 7)(x + 19) = 0
\(\Rightarrow\left\{\begin{matrix}x-7=0\\x+19=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=7\\x=-19\end{matrix}\right.\)
d. (x - 11)(x + 5) = 0
\(\Rightarrow\left\{\begin{matrix}x-11=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
2
Ta có: \(-\left(x-1\right)^2\le0\)
\(\Rightarrow2012-\left(x-1\right)^2\le2012\)
\(\Rightarrow M\le2012\)
Dấu '=' xảy ra khi x = 1
Vậy MMax = 2012 khi x = 1
3.
Ta có: \(\left|x-3\right|\ge0\)
\(\Rightarrow\left|x-3\right|+10\ge10\)
\(\Rightarrow N\ge10\)
Dấu '=' xảy ra khi x = 3
Vậy NMin = 10 khi x = 3
4.
\(n-6⋮n-4\)
\(\Rightarrow n-4-2⋮n-4\)
\(\Rightarrow-2⋮n-4\)
\(\Rightarrow n-4\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow n\in\left\{5;3;6;2\right\}\)
5.
\(n-5⋮n-2\)
\(\Rightarrow n-2-3⋮n-2\)
\(\Rightarrow-3⋮n-2\)
\(\Rightarrow n-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow n\in\left\{3;1;4;0\right\}\)
