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a) (2x-1)^3=27
b) (2x-1)^4=81
c) (x-2)^5=-32
d) (3x-1)^4=(3x-1)^6
đ) 5^x +5^x+2=650
g) 3^x-1 +5.3^x-1=162
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x-1)3 = 27
(2x-1)3 = 93
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2
b) (2x-1)4 = 81
(2x-1)4 = (\(\pm\)34)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5-\left|3x-1\right|=3\)
\(\left|3x-1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
\(\left|x+\frac{3}{4}\right|-5=-2\)
\(\left|x+\frac{3}{4}\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=3\\x+\frac{3}{4}=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=-\frac{15}{4}\end{cases}}\)
\(\left(1-2x\right)^2=9\)
\(\left(1-2x\right)^2=3^2\)
\(\Rightarrow1-2x=3\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-1\)
vậy \(x=-1\)
\(\left(x+5\right)^3=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
vậy \(x=-9\)
\(\left(2x+1\right)^2=\frac{4}{9}\)
\(\left(2x+1\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow2x+1=\frac{2}{3}\)
\(\Rightarrow2x=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{6}\)
vậy \(x=-\frac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
![](https://rs.olm.vn/images/avt/0.png?1311)
(-3/4)63x-1=(3/4)^3
3x-1=3+1
3x=3=1
x=4;3
x=4/3
Vậy x=4/3
![](https://rs.olm.vn/images/avt/0.png?1311)
tìm x biết
a) |2x−1|=x+4
* \(2x-1=x+4\)
\(<=> 2x-x=4+1\)
\(<=> x=5\)
* \(-2x-1=x+4\)
\(<=> -2x-x=4+1\)
\(<=> -3x=5\)
\(<=> x=\dfrac{-3}{5}\) (loại)
Vậy \(x=5\)
b)
\(<=> (3x-1)^4=3^4\)
\(<=> 3x-1=4\)
\(<=> 3x=5\)
\(<=> x=\dfrac{5}{3}\)
Vậy \(x=\dfrac{5}{3}\)
(3x−1)4=8c) (x−2)3=−64
\(<=> (x-2)^3=(-4)^3\)
\(<=> x-2=-4\)
\(<=> x=-2\)
Vậy \( x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a) \(|2x+1|-3=4x\)
\(\Leftrightarrow|2x+1|=4x+3\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=4x+3\\2x+1=-4x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-4x=3-1\\2x+4x=-3-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-2x=2\\6x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{-2}{3}\end{cases}}\)
a,
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\\ \)
\(\dfrac{1}{4}:x=\dfrac{8-15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
x = \(\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
b,
( 3x + 1)^3 = 64
(3x + 1)^3 = 4^3
(3x + 1) = 4
3x = 4 - 1
3x = 3
x = 3 : 3
x = 1
c,
( 2x - 3)^4 = 81
( 2x - 3) ^4 = 3^4
(2x - 3) = 3
2x = 3 + 3
2x = 6
x = 6: 2
x = 3