![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{x}{5}-\frac{2}{3}+2x=\frac{1}{2}\)
\(\frac{6x}{30}-\frac{20}{30}+\frac{60x}{30}=\frac{15}{30}\)
\(\frac{66x-20}{30}=\frac{15}{30}\)
\(\Rightarrow\) 66x - 20 = 15
66x = 15 + 20
66x = 35
x = \(\frac{35}{66}\)
Vậy x = \(\frac{35}{66}\)
b, \(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{7}\)
\(\frac{5}{2}-1+3x=\frac{1}{7}\)
\(\frac{3}{2}+3x=\frac{1}{7}\)
3x = \(\frac{1}{7}-\frac{3}{2}\)
3x = \(\frac{-19}{14}\)
x = \(\frac{-19}{42}\)
Vậy x = \(\frac{-19}{42}\)
Phần c lỗi, chắc như thế này
c, \(4\cdot\left(\frac{1}{2}-x\right)+\frac{1}{2}=\frac{-5}{6}+x\)
\(2-4x+\frac{1}{2}=\frac{-5}{6}+x\)
\(\frac{5}{2}-4x=\frac{-5}{6}+x\)
\(-4x-x=\frac{-5}{6}-\frac{5}{2}\)
\(-5x=\frac{-10}{3}\)
x = \(\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
Chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x - 1).3⁴ = 3⁷
2x - 1 = 3⁷ : 3⁴
2x - 1 = 3³
2x - 1 = 27
2x = 27 + 1
2x = 28
x = 28 : 2
x = 14
Câu b em ghi đề chính xác lại
Em xem câu b phải vầy không?
2ˣ⁺¹ + 4.2ˣ = 3.2¹²
2ˣ.(2 + 4) = 3.2.2¹¹
2ˣ.6 = 6.2¹¹
2ˣ = 2¹¹
x = 11
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2/3 + 1/3(2x-1) =-5
1/3(2x-1) = -5 - 2/3
1/3(2x-1) = -17/3
2x-1 = -17/3 : 1/3
2x-1 = -17
2x = -17 + 1
2x = -16
x = -16 : 2
x = -8
b) 25% .x + 2x = 4
x .( 25%+2 ) = 4
x . 9/4 = 4
x = 4 : 9/4
x = 16/9
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
a; - \(\dfrac{1}{3}\).(15\(x-9\)) + \(\dfrac{2}{7}\).(- \(x-34\)) = 1 - \(\dfrac{3}{4}\).(-16\(x+4\))
- 5\(x\) + 3 - \(\dfrac{2}{7}\)\(x\) - \(\dfrac{68}{7}\) = 1 + 12\(x\) - 3
12\(x\) + 5\(x\) + \(\dfrac{2}{7}x\) = 3 - \(\dfrac{68}{7}\) - 1 + 3
17\(x\) + \(\dfrac{2}{7}x\) = (3 - 1 + 3) - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\)\(x\) = 5 - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\) \(x\) = - \(\dfrac{33}{7}\)
\(x\) = - \(\dfrac{33}{7}\): \(\dfrac{121}{7}\)
\(x\) = - \(\dfrac{3}{11}\)
Vậy \(x\) = - \(\dfrac{3}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5^4:\left(2x-1\right)^2=25\\ \Leftrightarrow5^4:\left(2x-1\right)^2=5^2\\ \Leftrightarrow5^4:5^2=\left(2x-1\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^2=5^2\\\left(2x-1\right)^2=\left(-5\right)^2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b) \(17^3.x=17^4\\ \Rightarrow x=17^4:17^3=17\)
c) \(\left(2x+1\right)^3:49=7\\ \Leftrightarrow\left(2x-1\right)^3:7^2=7\\ \Leftrightarrow\left(2x-1\right)^3=7.7^2=7^3\\ \Leftrightarrow2x-1=7\\ \Leftrightarrow x=4\)
a)54 : (2x - 1)2 =25
54 : (2x - 1)2 =52
(2x - 1)2 =54 : 52
(2x - 1)2 = 52
2x - 1 = 5
2x = 5+1
2x = 6
x = 6: 2
x = 3
b)173.x = 174
x = 174 : 173
x = 17
c)(2x + 1)3 : 49 =7
(2x + 1)3 = 7 . 49
(2x + 1)3 = 343
(2x + 1)3 = 73
2x +1 = 7
2x = 7- 1
2x = 6
x = 6:2
x = 3
a) \(7-\left(2x-\dfrac{1}{3}\right)^2=3\)
\(\left(2x-\dfrac{1}{3}\right)^2=7-3\)
\(\left(2x-\dfrac{1}{3}\right)^2=4\)
\(\left(2x-\dfrac{1}{3}\right)^2=2^2\)
TH1: \(2x-\dfrac{1}{3}=2\)
\(2x=2+\dfrac{1}{3}=\dfrac{7}{3}\)
\(x=\dfrac{7}{3}:2=\dfrac{7}{6}\)
TH2: \(2x-\dfrac{1}{3}=-2\)
\(2x=-2+\dfrac{1}{3}=-\dfrac{5}{3}\)
\(x=-\dfrac{5}{3}:2=\dfrac{-5}{6}\)
b) \(\left(2x+\dfrac{1}{3}\right)^2+\dfrac{1}{2}=\dfrac{3}{4}\)
\(\left(2x+\dfrac{1}{3}\right)^2=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\left(2x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\left(2x+\dfrac{1}{3}\right)^2=\left(\dfrac{1}{2}\right)^2\)
TH1: \(2x+\dfrac{1}{3}=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}:2=\dfrac{1}{12}\)
TH2: \(2x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(2x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)
\(x=\dfrac{-5}{6}:2=\dfrac{-5}{12}\)