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0,2-|4,2-2x|=0
|4,2-2x|=0,2
=>4,2-2x=0,2
2x=4
x=2
=>4,2-2x=-0,2
2x=4,4
x=2,2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(0,2-\left|4,2-2x\right|=0\)
\(\Leftrightarrow\left|4,2-2x\right|=0,2\)
\(\Leftrightarrow\orbr{\begin{cases}4,2-2x=0,2\\4,2-2x=-0,2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2,2\end{cases}}\)
Vậy \(x\in\left\{2;2,2\right\}\)
\(0,2-\left|4,2-2x\right|=0\)
\(\Rightarrow\left|\frac{21}{5}-2x\right|=\frac{1}{5}\)
\(\Rightarrow\orbr{\begin{cases}\frac{21}{5}-2x=\frac{1}{5}\\\frac{21}{5}-2x=-\frac{1}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=\frac{22}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{11}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
0,2 - | 4,2 - 2x | = 0
| 4,2 - 2x | = 0,2 - 0
| 4,2 - 2x| = 0,2
=> 4,2 - 2x = 0,2 hoặc 4,2 - 2x = -0,2
+> 4,2 - 2x = 0,2
2x = 4,2 - 0,2
2x = 4
x = 4:2
x = 2
+> 4,2 -2x = -0,2
2x = 4,2+ 0,2
2x = 4,4
x= 4,4 :2
x = 2,2
0,2 - | 4,2 - 2x | = 0
| 4,2 - 2x | = 0,2 - 0
| 4,2 - 2x | = 0,2
=> 4,2 - 2x = \(\pm\)0,2
+) 4,2 - 2x = 0,2 +) 4,2 - 2x = - 0,2
2x = 4,2 - 0,2 2x = 4,2 - (- 0,2 )
2x = 4 2x = 4,4
x = 4:2 =2 x = 4,4:2 = 2,2
Vậy x \(\in\){ 2; 2,2}
![](https://rs.olm.vn/images/avt/0.png?1311)
(x-1)2=25
=>x-1=-5 hoặc x-1=5
=>x=-4 hoặc x=6
2x-1=5
2x=5+1=6
x=6:2
x=3
0,2-4,2-2x=0
-4-2x=0
2x=-4-0=-4
x=-4:2
x=-2
(x - 1)2 = 25
(x - 1)2 = 52
Suy ra x - 1 = 5
x = 5 + 1
x = 6. Vậy x = 6
2x - 1 = 5
2x = 5 + 1
2x = 6
x = 6 : 2
x = 3. Vậy x = 3
0,2 - 4,2 - 2x = 0
-4 - 2x = 0
2x = -4 - 0
2x = -4
x = -4 : 2
x = -2. Vậy x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\)
Vậy: \(x=3\) hoặc \(x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Ta có: \(\left|4,2-2.x\right|=0,2\)
\(\Rightarrow\orbr{\begin{cases}4,2-2x=0,2\\4,2-2x=-0,2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=4,4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2,2\end{cases}}\)
k cho mình nha cảm ơn
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(\dfrac{\left(-3\right)^x}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^x=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^x=-2187\)
\(\Leftrightarrow\left(-3\right)^x=\left(-3\right)^7\)
\(\Leftrightarrow x=7\)
Vậy ...
b/ \(2^{x-1}=16\)
\(\Leftrightarrow2^{x-1}=2^4\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=5\)
Vậy ....
c/ \(\left(x-1\right)^2=25\)
\(\Leftrightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy ....
d/ \(0,2-\left|4,2-2x\right|=0\)
\(\Leftrightarrow\left|4,2-2x\right|=0,2\)
\(\Leftrightarrow\left[{}\begin{matrix}4,2-2x=0,2\\4,2-2x=-0,2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=4,4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2,2\end{matrix}\right.\)
Vậy ............
e, \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
\(\Leftrightarrow\dfrac{5}{3}.\dfrac{4}{x}=20\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy ..
a) \(\dfrac{\left(-3\right)^x}{81}=-27\)
⇔ \(\dfrac{\left(-3\right)^x}{81}=\dfrac{-2187}{81}\)
⇔ (-3)x = -2187
⇔ (-3)x = (-3)7
⇔ x = 7
b) 2x-1 = 16
⇔ 2x-1 = 24
⇔ x - 1 = 4
⇔ x = 4 + 1
⇔ x = 5
c) (x - 1)2 = 25
⇔ \(\left[{}\begin{matrix}\left(x-1\right)^2=5^2\\\left(x-1\right)^2=\left(-5\right)^2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=5+1\\x=-5+1\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
d) 0,2 - |4,2 - 2x| = 0
⇔ |4,2 - 2x| = 0,2
⇔ \(\left[{}\begin{matrix}4,2-2x=0,2\\4,2-2x=-0,2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}2x=4,2-0,2\\2x=4,2-\left(-0,2\right)\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}2x=4\\2x=4,4\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=2\\x=2,2\end{matrix}\right.\)
e) \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
⇔ \(\dfrac{5}{3}.\dfrac{4}{x}=20\)
⇔ \(\dfrac{20}{3x}=20\)
⇔ 3x = 1
⇔ x = \(\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(4.3^{2x}-2.9^x-54=0\)
\(\Rightarrow3^{2x}\left(4-2\right)=54\)
\(\Rightarrow3^{2x}=27=3^3\)
Vì \(3\ne\pm1;3\ne0\) nên \(2x=3\Rightarrow x=\dfrac{3}{2}\)
b, \(\dfrac{1}{2}.2^x+4.2^x-288=0\)
\(\Rightarrow2^x\left(\dfrac{1}{2}+4\right)=288\)
\(\Rightarrow2^x=64=2^6\)
Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)