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H
3
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 7 2023
\(a,\\ \left(x+\dfrac{1}{2}\right)^3=\dfrac{8}{125}=\dfrac{2^3}{5^3}\\ \left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow\left(x+\dfrac{1}{2}\right)=\dfrac{2}{5}\\ x=\dfrac{2}{5}-\dfrac{1}{2}\\ x=-\dfrac{1}{10}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 7 2023
\(b,3\left|x\right|-27=\dfrac{1}{5}\\ 3\left|x\right|=\dfrac{1}{5}+27\\ 3\left|x\right|=\dfrac{136}{5}\\ \left|x\right|=\dfrac{136}{5}:3\\ \left|x\right|=\dfrac{136}{15}\\ Vậy:x=\dfrac{136}{15}.or.x=-\dfrac{136}{15}\)
SV
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 12 2022
a.\(\dfrac{1}{3}\) + x = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{1}{2}\)
b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\)
| x-1| = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)
|x-1| = \(\dfrac{3}{2}\)
\(\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}+1\\x=-\dfrac{3}{2}+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1
\(\dfrac{2}{3}\) (\(\dfrac{x}{2}\) + 3) = 1 - \(\dfrac{1}{3}\)
\(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)
\(\dfrac{x}{2}\) + 3 = 1
\(\dfrac{x}{2}\) = 1 - 3
\(\dfrac{x}{2}\) = -2
\(x\) = -4
d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)
(x+2)2 = 27.3
(x+2) =92
\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
OD
1
OY
7 tháng 8 2021
a) \(\dfrac{x.2}{-15}=\dfrac{-5}{3}\)
\(\dfrac{x.2}{-15}=\dfrac{25}{-15}\)
x.2=25
x=12,5
b) \(\dfrac{x-1}{-12}=\dfrac{-3}{x-1}\)
(x-1)2=-3.(-12)
(x-1)2=36
⇒(x-1)2\(\Rightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
7 tháng 8 2021
Lời giải:
$x+\frac{2}{-15}=\frac{-5}{3}$
$x=\frac{-5}{3}-\frac{2}{-15}=\frac{-5}{3}+\frac{2}{15}$
$x=\frac{-23}{15}$
18 tháng 5 2021
a) Ta có: \(\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x-1\right)\)
\(\Leftrightarrow2x-3-x+5=x+2-x+1\)
\(\Leftrightarrow x+2=3\)
hay x=1
Vậy: x=1
b) Ta có: \(2\left(x-1\right)-5\left(x+2\right)=-10\)
\(\Leftrightarrow2x-2-5x-10=-10\)
\(\Leftrightarrow-3x=-10+10+2=2\)
hay \(x=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
PT
18 tháng 5 2021
a, (2x - 3) - (x - 5) = (x + 2) - (x - 1)
2x - 3 - x + 5 = x + 2 - x + 1
(2x - x) + (-3 + 5) = (x - x) + (2 + 1)
x + 2 = 3
x = 1
12 tháng 9 2021
a) \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{3}{8}\)
b) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:0,02\\ \Rightarrow2\dfrac{2}{3}:x=\dfrac{800}{9}\\ \Rightarrow x=\dfrac{3}{100}\)
c) \(x^x-x+1=1\\ \Rightarrow x^x-x=0\\ \Rightarrow x^x=x\\ \Rightarrow x=1\)
d) \(5-\left|3x-1\right|=3\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
28 tháng 8 2019
\(\frac{1}{9}.3^4.3^x=3^7\)
\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
4 tháng 12 2021
a: \(\Leftrightarrow\left(x-1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
\(5^{x+2}=125\Rightarrow5^x.5^2=125\Rightarrow5^x=125:5^2\Rightarrow5^x=5\Rightarrow5^x=5^1\Rightarrow x=1\)
\(\left|x+3\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x+3\right|=\frac{1}{3}\)
\(\Rightarrow\begin{cases}x+3=\frac{1}{3}\\x+3=-\frac{1}{3}\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\)
a) 5x+2 = 125
=> 5x+2 = 52
=> x + 2 = 2
=> x = 2 - 2
=> x = 0
b) \(\left|x+3\right|-\frac{2}{3}=1\)
=> \(\left|x+3\right|=1+\frac{2}{3}\)
=> \(\left|x+3\right|=\frac{5}{3}\)
=> \(\left[\begin{array}{nghiempt}x+3=\frac{5}{3}\\x+3=-\frac{5}{3}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=-\frac{4}{3}\\x=-\frac{14}{3}\end{array}\right.\)