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a,(5x-1)6=36
5x-1=3
x=4/5
b,(2x+1)3=0,13
2x+1=0,1
x=-0,45
c,(2x-3)4=(2x-3)4(2x-3)2
(2x-3)2=0
2x-3=0
x=3/2
d,(2x+1)5=(2x+1)5(2x+1)2005
(2x+1)2005=0
2x+1=0
x=-1/2
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a)\(\orbr{\begin{cases}5x-1=3\\5x-1=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=4\\5x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{-2}{5}\end{cases}}\)
b) 2x+1=-0,1 <=> 2x=-1,1=>x=-0,55
c) (2x-3)4 .[1-(2x-3)2 ]=0
do (2x-3)4 lớn hơn 0 nên 1-(2x-3)2=0=>(2x-3)2=1=>2x-3=1=>2x=4=>x=2
d) tương tự câu c)
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Ta có: \(A=\left[6.\left(\frac{-1}{3}\right)^2-\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(\Rightarrow A=\left[6.\frac{1}{9}+\frac{1}{3}+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)
\(\Rightarrow A=\left[\frac{2}{3}+\frac{1}{3}+1\right]:\frac{-4}{3}\)
\(\Rightarrow A=\left[1+1\right].\frac{-3}{4}=2.\frac{-3}{4}=\frac{-3}{2}\)
Mà \(B=\left(729-1^3\right)\left(729-2^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...0...\left(729-125^3\right)=0\)
Vì \(\frac{-3}{2}< 0\)nên A < B
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a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55
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Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
Tìm x
a) ( 2x + 1 )3 = - 0,001
b) ( 2x + 1 )5 = ( 2x + 1 )2016
Giúp mk vs, mk ko wên ơn dou! Help me...
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a) (2x + 1)^3 = -0,001
(2x + 1)^3 = (-0,1)^3
=> 2x + 1 = -0,1
2x = -0,1 -1
2x = -1,1
x = (-1,1) : 2
x = -0,55
Vậy x = .......
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a,\(\left(5x-1\right)^6=729\)
\(\left(5x-1\right)^6=3^6\)
\(5x-1=3\)
\(5x=4\)
\(x=\dfrac{4}{5}\)
b Ta có \(8=2^3\),\(25=5^2\)
Mà \(\dfrac{8}{25}=\dfrac{2^x}{5^{x-1}}\)
=> \(2^3=2^x,5^2=5^{x-1}\)
=> x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
(5x - 1)(2x - 1/3) = 0
<=> 5x - 1 = 0 hoặc 2x - 1/3 = 0
=> x = 1/5 hoặc x = 1/6
vậy x= 1/5 hoặc x= 1/6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (5x - 1) . ( 2x - 1/3 ) = 0
=> 5x - 1 = 0
2x - 1/3 = 0
=> 5x = 1
2x = 1/3
=> x = 1/5
x = 1/6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(5x-1\right)\left(\frac{2x-1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}\)
b) \(6\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
a) \(\left(5x-1\right)\cdot\frac{2x-1}{3}=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\\frac{2x-1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=1\\2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=\frac{1}{5};x=\frac{1}{2}\)
b) 6(x-1)+2x(x-1)=0
<=> (x-1)(6+2x)=0
<=> \(\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy x=1; x=-3
a. \(\left(5x-1\right)^6=3^6\\ 5x-1=3\\ 5x=4\\ x=\dfrac{4}{5}\)
b. \(\left(2x+1\right)^3=\left(-0,1\right)^3\\ 2x+1=-0,1\\ 2x=-0,1-1\\ 2x=-1,1\\ x=-0,55\)
(5x-1)6 = 729
(5x-1)6 = 36
5x - 1 = - 3 hoặc 5x - 1 = 3
5x - 1 = -3
x = (-3 + 1) : 5 = -2/5
5x - 1 = 3
5x = 3 + 1
5x = 4
x = 4/5
vậy x ϵ {-2/5; 4/5}
b, (2x + 1)3 = -0,001 = (-0,1)3
2x + 1 = -0,1
2x + 1 = -0,1 - 1 = -1,1
x = -1,1 : 2
x = -0.55