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(3x-1)^6+746=810
(3x-1)^6=810-746
(3x-1)^6=64
(3x-1)^6=2^6
3x-1=2
3x=2+1
3x=3
x=3:3
x=1
Vậy x=1
chẳng biết đúng hay sai
(3x-1)^6+746=810
(3x-1)^6=64
(3x-1)^6=2^6
TH1:3x-1=2 TH2:3x-1=-2
3x=3 => x=1 3x=-1 => x=-1/3
Vậy x\(\in\) {-1;1}
Xin 1 k nha.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x - 10)37 = 0
2x - 10 = 0 : 37
2x - 10 = 0
2x = 0 + 10
2x = 10
x = 10 : 2
x = 5
b) 135(34 - x) = 810
34 - x = 810 : 135
34 - x = 6
x = 34 - 6
x = 28
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
3x + 3 - [7x+4] = 7 + [4x-1]
=> 3x + 3 - x - 4 = 7 + 4x - 1
=> 2x - 1 = 6 + 4x
=> 2x - 4x = 6 + 1
=> -2x = 7
=> x = -7/2
b,
3x+1 + 3x+3 =810
=> 3x+1[1 + 32] = 810
=> 3x+1 = 810 / 10
=> 3x+1 = 81
=> x = 4
c, \(1\frac{1}{2}:\left[\frac{1}{2}-\frac{1}{3}\right]-x=5\)
\(\Rightarrow\frac{3}{2}:\frac{1}{6}-x=5\Leftrightarrow9-x=5\)
\(\Leftrightarrow x=4\)
d,
\(2,4:\left[25\%+\frac{x}{40}\right]-\frac{12}{15}=3\frac{1}{5}\)
\(\Rightarrow\frac{12}{5}:\left[\frac{1}{4}+\frac{x}{40}\right]-\frac{12}{15}=\frac{16}{5}\)
\(\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=\frac{16}{5}+\frac{12}{15}\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=4\)
\(\Rightarrow\frac{10+x}{40}=\frac{12}{5}:4\Leftrightarrow\frac{10+x}{40}=\frac{3}{5}\)
\(\Rightarrow\frac{10+x}{40}=\frac{24}{40}\Leftrightarrow10+x=24\Rightarrow x=14\)
a) 3x + 3 - ( x + 4 ) = 7 + ( 4x - 1 )
3x + 3 - x - 4 = 7 + 4x - 1
2x - 1 = 6 + 4x
-2x = 7
\(\Rightarrow\)x = \(\frac{-7}{2}\)
b) 3x+1 + 3x+3 = 810
3x . 3 + 3x . 33 = 810
3x . ( 3 + 33 ) = 810
3x . 30 = 810
3x = 810 : 30
3x = 27
3x = 33
\(\Rightarrow\)x = 3
c) \(1\frac{1}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\frac{1}{6}-x=5\)
\(9-x=5\)
\(\Rightarrow x=9-5\)
\(\Rightarrow x=4\)
d) 2,4 : ( 25% + \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(3\frac{1}{5}\)
\(\frac{12}{5}\) : ( \(\frac{1}{4}\)+ \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(\frac{16}{5}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=\frac{16}{5}+\frac{12}{15}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{12}{5}:4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{3}{5}\)
\(\frac{x}{40}=\frac{3}{5}-\frac{1}{4}\)
\(\frac{x}{40}=\frac{7}{20}\)
\(\Rightarrow\frac{x}{40}=\frac{14}{40}\)
\(\Rightarrow x=14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Em kiểm tra lại đề bài, \(3^x+3^{x+2}+3^{x+2}\) hay \(3^x+3^{x+1}+3^{x+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có 2x + 2x + 1 + 2x + 2 = 56
⇒ 2x ( 1 + 21 + 22 ) = 56
⇒ 2x . 7 = 56
⇒ 2x = 56 : 7 = 8 = 23
Vậy x = 3
b) Ta có 3x + 3x + 2 + 3x + 3 = 111
⇒ 3x ( 1 + 32 + 33 ) = 111
⇒ 3x . 37 = 111
⇒ 3x = 111 : 37 = 3 = 31
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(3^x+3^{x+2}+3^{x+3}=3^5\cdot37\)
\(\Leftrightarrow3^x\cdot\left(1+9+27\right)=3^5\cdot37\)
hay x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(\Rightarrow x=2\)
a) 2.3x=162
=> 3x=162:2
=> 3x=81
=> 3x=34
=> x=4
b) 3x+3x+2=810
=> 3x+3x.32=810
=> 3x.(32+1)=810
=> 3x=810:10
=> 3x=81
=> 3x=34
=> x=4
2.3^x=162
3^x= 162:2
3^x= 81
3^x= 3^4
=> x=4