Bài 1:
a) 2|x-1| = 24.64
=> 2|x-1|= 210
=> |x-1|=10
=> \(\left[{}\begin{matrix}x-1=10\\x-1=-10\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-9\end{matrix}\right.\)
Vậy...
b)(3x-1)4=16
=> (3x-1)4=24
=> 3x - 1=2
=> 3x = 3
=> x=1
Vậy...
c) (2x+1)4=(2x+1)6
=> (2x+1)4 - (2x+1)6=0
=> (2x+1)4.[1 - (2x+1)2 ] = 0
=> \(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\1-\left(2x+1\right)^2=0\end{matrix}\right.\)
+) (2x+1)4=04
=> 2x+1=0
=> 2x = -1
=> x= \(\frac{-1}{2}\)
+) 1 - (2x+1)2=0
=> (2x+1)2 = 1
=> \(\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy...
d) x13=27.x10
=> x3=33
=> x=3
e)2x+2x+3=144
=> 2x(1+8)=144
=> 2x= 16 = 24
=> x=4
Bài 2:
a) Hình như đề bài là thế này:
CMR: 55-54+53 chia hết cho 7
Xét 55-54+53
=53(52-5+1)
=53. 21
Mà 21\(⋮\)7 => 53.21 chia hết cho 7 hay 55-54+53
Vậy...
b) Xét 76+75-74
= 74(72+7-1)
=74.55
Mà 55 \(⋮\)11 => 74.55 chia hết cho 11 hay 76+75-74 chia hết cho 7
Vậy...
Bài 1:
b) \(\left(3x-1\right)^4=16\)
=> \(\left(3x-1\right)^4=2^4\)
=> \(3x-1=2\)
=> \(3x=2+1\)
=> \(3x=3\)
=> \(x=3:3\)
=> \(x=1\)
Vậy \(x=1.\)
Bài 2:
\(5^5-5^4+5^3\)
\(=5^3.\left(5^2-5+1\right)\)
\(=5^3.21\)
Vì \(21⋮7\) nên \(5^3.21⋮7.\)
=> \(5^5-5^4+5^3⋮7\left(đpcm\right).\)
b) \(7^6+7^5-7^4\)
\(=7^4.\left(7^2+7-1\right)\)
\(=7^4.55\)
Vì \(55⋮11\) nên \(7^4.55⋮11\).
=> \(7^6+7^5-7^4⋮11\left(đpcm\right).\)
Chúc bạn học tốt!
