bài 1 tính
a) 2^2. 5^2-64 : 2^3
d) 3^5.9:243
c) 4.2^5. 2^3. \(\frac{1}{16}\)
d) \(\frac{11.3^{22}.3^7-9^{15}}{3^{28}}\)
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a) \(2^2\cdot5^2-64:2^3\)
\(=4\cdot25-2^6:2^3\)
\(=100-2^3=100-8\)
\(=92\)
b) \(3^5\cdot9:243\)
\(=3^5\cdot9:3^5=9\)
c) \(4\cdot2^5\cdot2^3\cdot\dfrac{1}{16}\)
\(=2^{10}\cdot\dfrac{1}{2^4}=2^6\)
\(=64\)
d) \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{3^{28}}\)
\(=\dfrac{11\cdot3^{29}-\left(3^2\right)^{15}}{3^{28}}\)
\(=\dfrac{11\cdot3^{29}-3^{30}}{3^{28}}=\dfrac{3^{29}\left(11-3\right)}{3^{28}}\)
\(=\dfrac{3^{29}\cdot8}{3^{28}}=3\cdot8=24\)
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{12}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}-\frac{1}{5}\)
\(B=\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
\(B=0-0+0+0-\frac{2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
\(B=\frac{-2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
Đến đây chỉ còn cách quy đồng thôi
A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
1/2^2 < 1/1*2
1/3^2 < 1/2*3
1/4^2 < 1/3*4
...
1/100^2 < 1/99*100
=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100
=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
=> A < 1 - 1/100
=> A < 1
minh deo can ban k dau :((
\(a,\frac{1}{2}x+\frac{3}{5}(x-2)=3\)
\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\Rightarrow\left[\frac{1}{2}+\frac{3}{5}\right]x=3+\frac{6}{5}\)
\(\Rightarrow\left[\frac{5}{10}+\frac{6}{10}\right]x=\frac{21}{5}\)
\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)
\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}=\frac{21}{5}\cdot\frac{10}{11}=\frac{21}{1}\cdot\frac{2}{11}=\frac{42}{11}\)
Vậy x = 42/11
a)
= 8/22 + 5/22
= 11/22
= 1/2
b)
= 24/8 - 3/8
= 21/8
c)
= 45/25
= 9/5
d)
= 3/7 x 1/4
= 3/28
:)
1^3-3^5-(-3^5)+1^64-2^9-1^36+1^15
=1+(-3^5+3^5)+1-2^9-1+1
=2-2^9
=-510
1: \(=\dfrac{2^{20}\cdot3^2+2^{24}}{2^{16}\cdot2^2\cdot5^2}=\dfrac{2^{20}\left(3^2+2^4\right)}{2^{18}\cdot5^2}=4\)
2: \(=\dfrac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
3: \(=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{2^2\cdot3^{28}}=2\cdot3=6\)