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Ta có : \(\frac{2x+5}{x+1}=\frac{2x+2+3}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=2+\frac{3}{x+1}\)
Vì 2 \(\inℤ\Rightarrow\frac{3}{x+1}\inℤ\Rightarrow3⋮x+1\Rightarrow x+1\inƯ\left(3\right)\Rightarrow x+1\in\left\{1;3;-1;-3\right\}\)
=> \(x\in\left\{0;2-2;-4\right\}\)
Để \(\frac{3x-1}{2x-1}\inℤ\Rightarrow3x-1⋮2x-1\Rightarrow2\left(3x-1\right)⋮2x-1\Rightarrow6x-2⋮2x-1\)
=> \(6x-3+1⋮2x-1\Rightarrow3\left(2x-1\right)+1⋮2x-1\)
Vì \(3\left(2x-1\right)⋮2x-1\)
=> \(1⋮2x-1\Rightarrow2x-1\inƯ\left(1\right)\Rightarrow2x-1\in\left\{1;-1\right\}\Rightarrow x\in\left\{1;0\right\}\)
\(\frac{2x+5}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=2+\frac{3}{x+1}\)
Để phân số nguyên => \(\frac{3}{x+1}\)nguyên
=> \(3⋮x+1\)
=> \(x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
=> \(x=\left\{0;-2;2;-4\right\}\)
\(\frac{3x-1}{2x-1}\)
Để phân số nguyên => \(3x-1⋮2x-1\)
=> \(2\left(3x-1\right)⋮2x-1\)
=> \(6x-2⋮2x-1\)
\(\Rightarrow3\left(2x-1\right)+1⋮2x-1\)
\(\Rightarrow1⋮2x-1\)
\(\Rightarrow2x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow x=\left\{1;0\right\}\)
\(a,2x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\forall Z\\x=1\end{cases}}}\)
\(b,x\left(2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
\(c;\left(x+1\right)+\left(x+3\right)+...............+\left(x+99\right)=0\)
\(\Rightarrow\left(x+x+...........+x\right)+\left(1+3+............+99\right)=0\)
\(\Rightarrow50x+2500=0\)
\(\Rightarrow50x=-2500\)
\(\Rightarrow x=-50\)
2/
\(a;\left(x-3\right)\left(2y+1\right)=7\)
\(\Rightarrow\left(x-3\right);\left(2y+1\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Xét bảng
| x-3 | 1 | -1 | 7 | -7 |
| 2y+1 | 7 | -7 | 1 | -1 |
| x | 4 | 2 | 10 | -4 |
| y | 3 | -4 | 0 | -1 |
Vậy...............................
\(b;xy+3x-2y=11\)
\(\Rightarrow x\left(y+3\right)-2y-6=11-6\)
\(\Rightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Rightarrow\left(x-2\right)\left(y+3\right)=5\)
\(\Rightarrow\left(x-2\right);\left(y+3\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét bảng'
| x-2 | 1 | -1 | 5 | -5 |
| y+3 | 5 | -5 | 1 | -1 |
| x | 3 | 1 | 7 | -3 |
| y | 2 | -8 | -2 | -4 |
Vậy................................
câu đầu bạn dưới làm rồi nên mình k làm lại
(2x+9)2=0
=> 2x+9=0
=> 2x=-9
=> x=-9/2
(2x-1)3=8
=> 2x-1=2
=> 2x=3
=> x=3/2
(1-3x)2=16
=> 1-3x=4
=> 3x=-3
=> x=-1
(3x+1)+1=-26
=> 3x=-27
=> x=-9
(x+1)+(x+3)+(x+5)+...+(x+2017)=0
(x+x+x+...+x)+(1+3+5+...+2017)=0
=> 1009x+1018081=0
1009x=-1018081
=> x=-1009
1,Tim x:
a,2x-18=10
b,3x-26=-5
c,/x-2/-15=-9
d,2x+11=x-4
Ai làm nhanh nhất mình tick cho.Cảm ơn trước nha
a) 2x- 18 = 10
2x = 28
x = 14
b) 3x - 26 = -5
3x = 21
x = 7
c) |x - 2| - 15 = |-9|
|x - 2\ = 6
x - 2 = 6 => x= 8
x - 2 = -6 => x = -4
d) 2x+ 11 = x- 4
x - 2x = 11 + 4
-x = 15
x = -15
\(\dfrac{3}{5}\) x 0,25\(x\) = - \(\dfrac{1}{2}\)
0,25\(x\) = - \(\dfrac{1}{2}\) : \(\dfrac{3}{5}\)
0,25\(x\) = - \(\dfrac{1}{2}\) x \(\dfrac{5}{3}\)
0,25\(x\) = - \(\dfrac{5}{6}\)
\(x\) = - \(\dfrac{5}{6}\) : 0,25
\(x\) = - \(\dfrac{5}{6}\) x 4
\(x\) = - \(\dfrac{10}{3}\)
Vậy \(x\) = - \(\dfrac{10}{3}\)
2626 : (\(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\)) = 26
\(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\) = 2626 : 26
\(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\) = 101
\(x\) x ( \(\dfrac{1}{2}\) + \(\dfrac{5}{2}\)) = 101
\(x\) x 3 = 101
\(x\) = 101 : 3
\(x\) = \(\dfrac{101}{3}\)
Vậy \(x\) = \(\dfrac{101}{3}\)