Bài 1 a 2^19 * 27^3/ 4^9. 3
b C= 1+3+3^2+3^3+....+ 3^99
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TC
8 tháng 5 2022
a)\(=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
b)\(=\dfrac{5}{9}\times\dfrac{3}{2}=\dfrac{15}{18}=\dfrac{5}{6}\)
d)\(=\left(\dfrac{12}{8}-\dfrac{3}{8}\right)\times2=\dfrac{9}{8}\times2=\dfrac{18}{8}=\dfrac{9}{4}\)
c)\(=\dfrac{4}{3}-\dfrac{5}{6}=\dfrac{8}{6}-\dfrac{5}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
Q
8 tháng 5 2022
a) 1 + 4/3 = 7/3
b) 5/9 : 2/3 = 5/6
c ) 4/3 -1/3 x 5/2
= 1 x 5/2
= 5/2
d) ( 3/2 - 3/8) : 1/2
= 9/8 : 1/2
= 9/4
e) 15/16 : 3/8 x 3/4
= 5/2 x 3/4
= 15/8
f) 7/19 x 1/3 x 7/19 x 2/3
= 7/19 x (1/3 x 2/3)
= 7/19 x 2/9
= 14/171
g) 3/5 x 8/27 x 25/3
= 3/5 x 25/3 x 8/27
= 5 x 8/27
= 40/27
h) 1/5 + 4/11 + 4/5 + 7/11
= (1/5 + 4/5) + (4/11 + 7/11)
= 1 + 1
= 2
30 tháng 8 2020
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
19 tháng 6 2021
ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều
31 tháng 1 2019
Bài 3 :
a) \(1+\left(-2\right)+3+\left(-4\right)+...+19+\left(-20\right)\)\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[19+\left(-20\right)\right]\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right)\cdot10=-10\)
b) \(1-2+3-4+...+99-100=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right)\cdot50=-50\)
c) \(2-4+6-8+...+46-48+50-52=\left(2-4\right)+\left(6-8\right)+...+\left(50-52\right)\)
\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)\)
\(=\left(-2\right)\cdot13=-26\)
d) \(-1+3-5+7-...-97+99\)\(=\left(-1+3\right)+\left(-5+7\right)+...+\left(-97+99\right)\)
\(=2+2+...+2\)
\(=2\cdot25=50\)
e) \(1+\left(-2\right)+3+\left(-4\right)+...+1999+\left(-2000\right)+2001\)\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[1999+\left(-2000\right)\right]+2001\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2001\)
\(=\left(-1\right)\cdot1000+2001=-1000+2001=1001\)
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Bài 4 :
a) \(\left(2ab^2\right):\left(abc\right)=\left[2\cdot4\cdot\left(-6^2\right)\right]:\left[4\cdot\left(-6\right)\cdot12\right]\)
\(=\left[2\cdot4\cdot36\right]:\left[4\cdot\left(-6\right)\cdot12\right]\)
\(=\left[8\cdot36\right]:\left[-24\cdot12\right]\)
\(=288:\left(-288\right)=-1\)
b) \(\left[\left(-25\right)\cdot\left(-27\right)\cdot\left(-x\right)\right]:y=\left[\left(-25\right)\cdot\left(-27\right)\cdot4\right]:\left(-9\right)\)
\(=\left[675\cdot4\right]:\left(-9\right)=2700:\left(-9\right)=-300\)
c) \(\left(a^2-b^2\right):\left(a+b\right)\left(a-b\right)=\left(5^2-\left(-3^2\right)\right):\left(5+\left(-3\right)\right)\left(5-\left(-3\right)\right)\)
\(=\left(25-9\right):\left(5+\left(-3\right)\right)\left(5-\left(-3\right)\right)\)
\(=16:2\cdot8=8\cdot8=64\)
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
12 tháng 8 2018
a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)
b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
\(\frac{2^{19}\cdot27^3}{4^9\cdot3}=\frac{2^{19}\cdot\left(3^3\right)^3}{\left(2^2\right)^9\cdot3}=\frac{2^{19}\cdot3^9}{2^{18}\cdot3}=\frac{2\cdot3^8}{1\cdot1}=2\cdot3^8\)
\(C=1+3+3^2+3^3+....+3^{99}\)
\(\Rightarrow3C=3+3^2+3^3+3^4+...+3^{100}\)
\(\Rightarrow3C-C=\left(3+3^2+3^3+3^4+...+3^{100}\right)-\left(1+3+3^2+3^3+....+3^{99}\right)\)
\(\Rightarrow2C=3^{100}-1\)
\(C=\frac{3^{100}-1}{2}\)