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Ta có \(\left(x-5\right)^4=\left(x-5\right)^6\)
=> \(\orbr{\begin{cases}x-5=\pm1\\x-5=0\end{cases}}\)
(+) \(x-5=-1=>x=4\)
(+) \(x-5=1=>x=6\)
(+) \(x-5=0=>x=5\)
Vậy
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\orbr{\begin{cases}x-5=\pm1\\x-5=0\end{cases}}\)
+) x - 5 = 1 \(\Rightarrow\)x = 6
+) x - 5 = -1 \(\Rightarrow\)x = 4
+) x - 5 = 0 \(\Rightarrow\)x = 5
Vậy x \(\in\){6;4;5}
![](https://rs.olm.vn/images/avt/0.png?1311)
5-[-|-4|+(-3+x-|-6|)]=15-94+|-8|-9
5-[-4+(-3+x-6)] = 15-94+8-9
5-[-4+(-3)+x-6] = (-79)+8-9
5+4-(-3) - x + 6 = -71 - 9
9-(-3) - x + 6 = -80
12 - x + 6 = -80
12-x = -80 - 6
12-x = -86
x = 12 - (-86)
x = 98
mk ko chắc lắm
5 - [ - l -4 l + ( - 3 + x - l - 6 l ) ] = 15 - 94 + l -8 l - 9
5 - [ - 4 + ( - 3 + x - 6 ) ] = - 78 + 8 - 9
5 - [ - 4 - 3 + x - 6 ] = - 70 - 9
5 + 4 + 3 - x + 6 = - 79
9 + 3 - x + 6 = - 79
12 - x + 6 = - 79
12 + ( - x ) + 6 = - 79
12 + ( - x ) = - 79 - 6
12 + ( - x ) = - 85
- x = - 85 - 12
- x = - 97 => x = 97
Vậy x = 97
![](https://rs.olm.vn/images/avt/0.png?1311)
5/8 + 0,2 : |-x/4 + 2/5| = 3/2
0,2 : |-x/4 + 2/5| = 3/2 - 5/8
0,2 : |-x/4 + 2/5| = 7/8
|-x/4 + 2/5| = 7/8 . 0,2
|-x/4 + 2/5| = 7/40
\(\Rightarrow\orbr{\begin{cases}-\frac{x}{4}+\frac{2}{5}=\frac{7}{40}\\-\frac{x}{4}+\frac{2}{5}=-\frac{7}{40}\end{cases}\Rightarrow\orbr{\begin{cases}-\frac{x}{4}=-\frac{9}{40}\\-\frac{x}{4}=-\frac{23}{40}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{10}\\x=\frac{23}{10}\end{cases}}}\)
Báo cáo sai phạm
5/8 + 0,2 : |-x/4 + 2/5| = 3/2
0,2 : |-x/4 + 2/5| = 3/2 - 5/8
0,2 : |-x/4 + 2/5| = 7/8
|-x/4 + 2/5| = 7/8 . 0,2
|-x/4 + 2/5| = 7/40
⇒[
−x4 +25 =7/40 |
−x4 +25 =−7/40 |
⇒[
−x4 =−9/40 |
−x4 =−23/40 |
⇒[
x=9/10 |
x=2/310 |
![](https://rs.olm.vn/images/avt/0.png?1311)
-12(x-5) + 7(3-x) = 5
-12x + 60 + 21 - 7x = 5
-12x + (-7x) = 5 + (-21) + (-60)
-19x = -76
19x = 76
x = 76 : 19
x = 4
k mk nha bn
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\left(x-5\right)^4-\left(x-5\right)^4\cdot\left(x-5\right)^2=0\)
\(\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\1-\left(x-5\right)^2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x-5=\left\{\pm1\right\}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=\left\{6;4\right\}\end{cases}}\)
Vậy \(x=\left\{4;5;6\right\}\)
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-6\right)\left(x-4\right)=0\end{cases}}\)
1: \(x-5=0\Leftrightarrow x=5\)
2.\(\left(x-6\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
Vậy ...