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bài 1) thực hiện phép tính
a) \(4^5-81:3^2=1024-9=1015\)
b) \(3^2.22-3^2.12=3^2.2\left(11-6\right)=18.5=90\)
c) \(2^3.15-\left[115-\left(12-5\right)\right]=120-\left(115-12+5\right)=120-115+12-5=12\)
d) \(3.3^2-19^{21}:19^{20}+2010^0=27-19+1=9\)
e)\(7^{25}:\left(7^{21}.46+7^{21}.3\right)=7^{25}:\left(7^{21}.49\right)=\frac{7^{21}.7^4}{7^{21}.49}=\frac{2401}{49}=49\)
bài 2)Tim x
a) 716 - (x - 143) = 695
<=> x - 143 = 716 - 695
<=> x - 143 = 21
<=> x = 21 + 143
<=> x = 164
vậy x = 164
1. 53 = 5.5.5 = 125
2. 27 = 2.2.2.2.2.2.2 = 128
3. 44 = 4.4.4.4 = 256
4. 73 = 7.7.7 = 343
6. 35 = 243
7. 26 = 64
8. 34 = 81
9. 83 = 512
11. 132 = 169
12. 112 = 121
13. 142 = 196
14. 152 = 225
16. 172 = 289
17. 182 = 324
18. 192 = 361
19. 202 = 400
21. 104 = 10000
22. 105 = 100000
23. 106 = 1000000
24. 107 = 10000000
Thế này à:
a, \(\frac{2^{13}+25}{2^{10}}\)
b, \(\frac{21^2.14.125}{35^5.6}\)
c,\(\frac{45^3.20^4.18^2}{180^5}\)
d, \(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
2 mũ 13 ta chia cho 2 mũ 10 ta có là 2 mũ 3 chuyển thành 8 rồi cộng cho 25 ta có là 33
a)64:2mũ5×30×4
= 64 : 32 x 30 x 4
= 240
b)3 mũ 2× 5 - 2 mũ 2×7+2 mũ 0 × 5
= 9 x 5 - 4 x 7 + 1 x 5
= 45 - 28 + 5
= 22
c)2 mũ 3-5 mũ 3÷5 mũ 2 + 12×2 mũ 2
= 8 - 125 : 25 + 12 x 4
= 8 - 5 + 48
= 51
d)2[(7-3 mũ 3÷3 mũ 2) chia 2 mũ 2 + 99]-100
= 2[( 7 - 27 : 9) : 4 + 99] - 100
= 2[4 : 4 + 99] - 100
= 2. 100 - 100
= 200 - 100
= 100
e)4[(3 + 3^7:3^4)chia 10 + 97]-300
= 4[( 3 + 3^3) : 10 + 97] - 300
= 4[ 30 : 10 + 97 ] - 300
= 4. 100 - 300
= 400 - 300
= 100
f)2^2 x 5 [(5 mũ 2 cộng 2 mũ 3) chia 11 - 2] - 3^2 x 2
= 4 x 5 [ (25 + 8 ) : 11 - 2] - 9 x 2
= 20 [ 33 : 11 - 2] - 18
= 20. 1 - 18
= 20 - 18
= 2
a,\(5^3.2-100:4+2^3.5\)
= 125 . 2 - 25 + 8 . 5
= 250 - 25 + 40
= 265
b, \(6^2:9+50.2-3^3.3\)
= 36 : 9 + 100 - 27 . 3
= 4 + 100 - 81
= 23
Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
a) 3 . 52 + 15 . 22 - 26 : 2
= 3 . 25 + 15 . 4 - 26 : 2
= 75 + 60 - 13
= 135 - 13
= 122
b) 53 . 2 - 100 : 4 + 23 . 5
= 125 . 2 - 100 : 4 + 8 . 5
= 250 - 25 + 40
= 225 + 40
= 265
c) (519 : 517 + 3 ) : 7
= (52 + 3) : 7
= (25 + 3) : 7
= 28 : 7
= 4
d) 79: 77 - 32 + 23 . 52
= 72 - 32 + 23 . 52
= 49 - 9 + 8 . 25
= 40 + 200
= 240
e) 791 : 789 + 5 . 52 - 124
= 72 + 5 . 25 - 124
= 49 + 125 - 124
= 174 - 124
= 50
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13
