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6 tháng 10 2023
`#3107.101107`
a)
\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)
Vậy, `x = 4`
b)
\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)
Vậy, `x = 3`
c)
\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
Vậy, `x = 4.`
ND
Nguyễn Đức Trí
VIP
7 tháng 10 2023
a) \(27< 3^x< 243\)
\(\Rightarrow3^3< 3^x< 3^5\)
\(\Rightarrow3< x< 5\)
c) \(3^x+3^{x+2}=810\)
\(\Rightarrow3^x\left(1+3^2\right)=810\)
\(\Rightarrow3^x.10=810\)
\(\Rightarrow3^x=810:10\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
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
\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
NT
0
TH
3
NC
1
LD
2 tháng 1 2022
27:(x-3/2)^3=(x-3/2):3
Ta có: \(\dfrac{27}{\left(x-\dfrac{3}{2}\right)^3}=\dfrac{\left(x-\dfrac{3}{2}\right)}{3}\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^3.\left(x-\dfrac{3}{2}\right)\)=27.3
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4\)=81
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}+\dfrac{3}{2}\\x=\dfrac{-8}{2}+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy x∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)
4 tháng 10 2021
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
PT
0
XO
4 tháng 8 2020
Ta có\(\frac{2-x}{-3}=\frac{27}{x-2}\)
(2 - x)(x - 2) = 27.(-3)
=> -(x - 2)2 = - 81
=> (x - 2)2 = 81
=> \(\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-7\end{cases}}\)
4 tháng 8 2020
\(\frac{2-x}{-3}=\frac{27}{x-2}\)
\(\Leftrightarrow\left(2-x\right)\left(x-2\right)=-3\cdot27\)
\(\Leftrightarrow-\left(x-2\right)\left(x-2\right)=-81\)
\(\Leftrightarrow-\left(x-2\right)^2=-81\)
\(\Leftrightarrow\left(x-2\right)^2=81\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-7\end{cases}}\)
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
\(b,3\left|x\right|-27=\dfrac{1}{5}\\ 3\left|x\right|=\dfrac{1}{5}+27\\ 3\left|x\right|=\dfrac{136}{5}\\ \left|x\right|=\dfrac{136}{5}:3\\ \left|x\right|=\dfrac{136}{15}\\ Vậy:x=\dfrac{136}{15}.or.x=-\dfrac{136}{15}\)
23 tháng 5 2016
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
ML
23 tháng 5 2016
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
--->( 3x -2)^2 = 25 = 5^2
---> ( 3x - 2) = 5
---> 3x= 7
---> x = 7/3
Vậy x = 7/3
TL:
Vậy x = 7/3
-HT-