Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
a) x- 18 = - 25
x =( -25 ) + 18
x = -7
b) 42 - x = 53 - 60
16 - x = 65
x = 16 - 65
x = -49
a) 33x+1 . 5 = 10935
=> 33x . 3 . 5 = 10935
=> 27x . 3 . 5 = 10935
=> 27x = 729
=> 27x = 272
=> x = 2
b) 25x + 1 . 3 = 6144
=> 25x . 2 . 3 = 6144
=> 32x . 6 = 6144
=> 32x = 1024
=> 32x = 322
=> x = 2
c) (2x + 3)4 = 625
=> (2x + 3)4 = (\(\pm\)5)4
=> \(\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
d) (3x + 1)3 = 343
=> (3x + 1)3 = 73
=> 3x + 1 = 7 => 3x = 6 => x = 2
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
Giải:
a) \(4\left(x+1\right)-\left(3x+1\right)=14\)
\(\Leftrightarrow4x+4-3x-1=14\)
\(\Leftrightarrow x+3=14\)
\(\Leftrightarrow x=14-3=11\)
Vậy ...
b) \(-2\left(x-1\right)-\left(2-3x\right)=2\left(x+1\right)-4\)
\(\Leftrightarrow-2x+2-2+3x=2x+2-4\)
\(\Leftrightarrow x=2x-2\)
\(\Leftrightarrow x-2x=-2\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\)
Vậy ...
c) \(-\left(x-3\right)+\left(4x-1\right)-14=2\left(x-2\right)-5\)
\(\Leftrightarrow-x+3+4x-1-14=2x-4-5\)
\(\Leftrightarrow3x-15=2x-9\)
\(\Leftrightarrow3x-2x=-9+15\)
\(\Leftrightarrow x=6\)
Vậy ...
\(1/3x=5/4+3/4=>1/3x=8/4=>x=2:1/3=2*3=>x=6\)
\(\dfrac{1}{3}x-\dfrac{3}{4}=\dfrac{5}{4}\\ \dfrac{1}{3}x=\dfrac{5}{4}+\dfrac{3}{4}\\ \dfrac{1}{3}x=2\\ x=2:\dfrac{1}{3}\\ x=6\)