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Bài 1:
a) \(x^{10}=1^x\Rightarrow\orbr{\begin{cases}x=1\\x=10\end{cases}}\)
b) \(x^{10}=x\Rightarrow x=1\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5.\left(2x-15\right)^3=\left(2x-15\right)^3\)
\(\left(2x-15\right)^2=1\Rightarrow x=8\)
Bài 2:
\(a;2^{16}=2^{13}\cdot2^3=2^{13}\cdot8>7\cdot2^{13}\)
\(b;49^8\cdot27^5=7^{16}\cdot3^{15}=21^{15}\cdot7>21^5\)
C;Ta có:\(199^{20}< 200^{20}=2^{20}\cdot10^{40}=2^{15}\cdot10^{40}\cdot2^5\)
\(2003^{15}>2000^{15}=2^{15}\cdot10^{45}=2^{15}\cdot10^{40}\cdot10^5\)
Vì 25<105 nên 19920<200315
\(d;3^{39}< 3^{40}=9^{20}< 11^{20}< 11^{21}\)
2x : 32 = 128
=> 2x = 128.32
=> 2x = 4096
=> 2x = 212
=> x = 12
X=12 X=5 X k thỏa mãn X=5 X=7 X k thỏa mãn X=1 lần lượt theo thứ tự nha
- Bài 1:
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.13+2^{10}.13.5}{2^8.2^2.13.2}\)
\(=\frac{2^{10}.13\left(1+5\right)}{2^{10}.13.2}=\frac{2^{10}.13.6}{2^{10}.13.2}=\frac{6}{2}=3\)
\(B=\left(1+2+3+...+100\right)\left(1^2+2^2+3^2+...+100^2\right)\left(65.111-13.15.37\right)\)
\(=\left(1+2+3+...+100\right)\left(1^2+2^2+...+100^2\right)\left(65.111-13.5.3.37\right)\)
\(=\left(1+2+...+100\right)\left(1^2+2^2+...+100^2\right)\left(65.111-65.111\right)\)
\(=\left(1+2+...+100\right)\left(1^2+2^2+...+100^2\right).0\)
\(=0\)
- Bài 2:
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(x+1+x+2+x+3+...+x+100=5750\)
\(x+x+x+...+x+1+2+3+...+100=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(x=700:100\)
\(x=7\)
t_i_c_k cho mình nha ^^
\(a,[\left(8.x-12\right):4].3^3.3=3^6.6\)
\(\left(8x-12\right):4=54\)
\(8x-12=216\)
\(8x=228\)
\(x=28,5\)
\(b,41-2^{x+1}=9\)
\(2^{x+1}=41-9\)
\(2^{x+1}=32\)
\(2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
1) [(52.23-72.2):2].6-7.2
=[(25.8-49.2):2].6-49
=[(200-98):2].6-49
=[102:2].6-49
=51.6-49
=306-49
=257
2)15.(27+18+6)+15.(23+12)
=15.51+15.35
=15.(51+35)
=15.86
=1290
3)(81:33.15):9.(40:8.2)
=(81:27.15):9 .(5.2)
=(3.15):9.10
=45:9.10
=5.10
=50
\(5x+2x=6^2-5^0\Leftrightarrow5x+2x=6^2-1\)
\(\Leftrightarrow5x-2x=36-1\Leftrightarrow5x-2x=35\Leftrightarrow x\left(5+2\right)=35\)
\(\Leftrightarrow7x=35\Leftrightarrow x=35:7\Leftrightarrow x=5\)
\(5x+x=150:2+3\Leftrightarrow5x+x=75+3\)
\(\Leftrightarrow5x+x=78\Leftrightarrow x\left(5+1\right)=78\Leftrightarrow x6=78\)
\(\Leftrightarrow x=78:6\Leftrightarrow x=13\)
a) 5x + 2x = 62 - 50
7x = 35 => x = 5
b) 5x + x = 150 : 2 + 3
6x = 78 => x = 13
c) 6x + x = 511 : 59 + 31
7x = 28 => x = 4
d) 5x + 3x = 36 : 33 x 4 + 12
8x = 120 => x = 15
e) 4x + 2x = 68 - 219 : 216
6x = 60 => x = 10
f) 5x + x = 39 - 311 : 39
6x = 30 => x = 5
g) 7x - x = 521 : 519 + 3 x 22 - 70
6x = 36 => x = 6
h) 7x - 2x = 617 : 615 + 44 : 11
5x = 40 => x = 8
a, A = 1 + 2 + 22 + 23 +...+ 239
A = 1.(1 + 2 + 22 + 23) + 24.(1+2+22 + 23)+...+236(1+2+22+23)
A = 1.15 + 24.15 +...+ 236.15
A = (1 +24 +...+236).15 chia hết 15 ( do 15 chia hết 15)
Vậy A chia hết 15.
b, T = 1257 - 259
T = (53)7 - (52)9
T = 521 - 518
T = 518(53 - 1)
T = 518.124 chia hết 124 ( do 124 chia hết 124)
Vậy T chia hết 124
(7x-11)3=25.52+200
=> (7x-11)3=800+200
=> (7x-11)3=1000
=> (7x-11)3=103
=> 7x - 11 = 10
=> 7x = 21
=> x = 3
(2x-15)5=(2x-15)3
=> (2x-15)5 - (2x-15)3 = 0
=> (2x-15)3 . [ (2x-15)2 - 1 ] = 0
=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}2x=15\\2x=16\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)
Mà x thuộc N
=> x = 8
(3x-5)10=(3x-5)9
=> (3x-5)10 - (3x-5)9 = 0
=> (3x-5)9 .[ (3x-5) - 1 ] = 0
=> \(\orbr{\begin{cases}\left(3x-5\right)^9=0\\\left(3x-5\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}3x-5=0\\3x-5=1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=5\\3x=6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
Mà x thuộc N
=> x = 2
a)(7x-11)^3=1000
(7x-11)^3=10^3
7x-11 =10
7x =10+11=21
x =21:7=3