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bài 1:
a) 1/3x=-5/3
x=-5
b) x+12/5=37/15
x=1/15
c) x-1/7=-36/7
x=-5
d) 3x-1/2=0
3x=1/2
x=1/6
e) 2/5+x=1/4
x=-3/20

\(a,2.\left(x+3\right)-3x=-2x+7\)
\(\Leftrightarrow2x+6-3x=-2x-7\)
\(\Leftrightarrow-x+6=-2x+7\)
\(\Leftrightarrow x=-6+7\)
\(\Leftrightarrow x=1\)
\(b,5.\left(x-3\right)-29=-2.\left(7-x\right)-23\)
\(\Leftrightarrow5x-15-29+23=14-2x\)
\(\Leftrightarrow5x-21=14-2x\)
\(\Leftrightarrow5x+2x=35\)
\(\Leftrightarrow x=5\)
\(c,-17+\left|5-x\right|=-2.7\)
\(\Leftrightarrow-17+\left|5-x\right|=-14\)
\(\Leftrightarrow\left|5-x\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=3\\5-x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=8\end{cases}}\)
\(d,21-\left|x+7\right|=-42:\left(-2\right)\)
\(\Leftrightarrow21-\left|x+7\right|=21\)
\(\Leftrightarrow\left|x+7\right|=0\)
\(\Leftrightarrow x+7=0\)
\(\Leftrightarrow x=7\)
\(f,\left|3x-5\right|-\left(-15\right)=5\)
\(\Leftrightarrow\left|3x-5\right|+15=5\)
\(\Leftrightarrow\left|3x-5\right|=-10\)
Vì \(\left|a\right|\ge0\)mà \(-10< 0\)nên \(x\in\Phi\)
Đã giải quyết xong!!!
giúp mình với: so sánh -37/56 với -377/567 ai giải đc mình cho

a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)

bài 1:
a) ta có: 3x + 5 = (3(x+1)+2)\(⋮\)(x+1)
vì (3(x+1)\(⋮\)(x+1) nên 2 \(⋮\)(x+1) => (x+1) \(\in\)Ư(2) => (x+1) \(\in\)\(\xi\)-2;-1;1;2 \(\xi\)=> x \(\in\)\(\xi\)-3; -2; 0; 1 \(\xi\)
vậy, x= -3; -2; 0; 1

a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)

(3x-1):5=7
(3x-1)=7x5
3x-1=35
3x=35+1
3x=36
x=36:3
x=12