tính:
S=3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)
o) \(\dfrac{\left(-1\right)^6.3^5.4^3}{9^2.2^5}\)
s) \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
t) \(\sqrt{\dfrac{4}{9}}\) - \(\dfrac{1}{2}\): \(\left|\dfrac{-2}{3}\right|\)
Mg giải gấp giúp mình với
o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)
s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)
\(=2:\dfrac{3}{2}=\dfrac{4}{3}\)
t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)
\(=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}\)
\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)
S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}...+\dfrac{1}{9.10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{2}{5}\)
S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9}< 1-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{8}{9}\)
Vậy \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
Bài 1. Tính
A= \(\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
B= \(\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
Bài 2. Tính
a) \(5\dfrac{1}{2}.3\dfrac{1}{4}\) b) \(6\dfrac{1}{3}:4\dfrac{2}{9}\) c) \(4\dfrac{3}{7}.2\)
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
Bài 2:
a) \(5\dfrac{1}{2}.3\dfrac{1}{4}=\dfrac{11}{2}.\dfrac{13}{4}=\dfrac{11.13}{2.4}=\dfrac{143}{8}\)
b) \(6\dfrac{1}{3}:4\dfrac{2}{9}=\dfrac{19}{3}:\dfrac{38}{9}=\dfrac{19}{3}.\dfrac{9}{38}=\dfrac{3}{2}\)
c) \(4\dfrac{3}{7}.2=\dfrac{31}{7}.2=\dfrac{31.2}{7}=\dfrac{62}{7}\)
1. Tính
a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)
b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)
c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)
Giúp tui với mình cần gấp lắm.
a)
`2/3+5/2-3/4`
`=10/4-3/4+2/3`
`=7/4+2/3`
`=21/12+8/12`
`=29/12`
b)
`2/5xx1/2:1/3`
`=2/10xx3/1`
`=6/10=3/5`
c)
`2/9:2/9xx1/3`
`=2/9xx9/2xx1/3`
`=1xx1/3`
`=1/3`
a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)
= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)
= \(\dfrac{38-9}{12}\)
= \(\dfrac{29}{12}\)
b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)
= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)
= \(\dfrac{3}{5}\)
c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)
= 1 x \(\dfrac{1}{3}\)
= \(\dfrac{1}{3}\)
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
1) tính \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{19683}\)
2) tính \(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{78125}\)
1: \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^9}\)
\(=\left(\dfrac{1}{3}\right)^0+\left(\dfrac{1}{3}\right)^1+...+\left(\dfrac{1}{3}\right)^9\)
u1=1; q=1/3
\(S_9=\dfrac{u1\cdot\left(1-q^9\right)}{1-q}=\dfrac{1\left(1-\left(\dfrac{1}{3}\right)^9\right)}{1-\dfrac{1}{3}}\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3^9}\right)\)
2:
\(S=\left(\dfrac{1}{5}\right)^0+\left(\dfrac{1}{5}\right)^1+...+\left(\dfrac{1}{5}\right)^7\)
\(u1=1;q=\dfrac{1}{5}\)
\(S_7=\dfrac{1\cdot\left(1-q^7\right)}{1-q}=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}=\dfrac{5}{4}\left(1-\dfrac{1}{5^7}\right)\)
1) tính \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{19683}\)
2) tính \(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{78125}\)
1, Ta có \(\dfrac{\dfrac{1}{3}}{1}=\dfrac{1}{3};\dfrac{\dfrac{1}{9}}{\dfrac{1}{3}}=\dfrac{1}{3};...\)
-> Là cấp số nhân, q = 1/3
Ta có \(S_9=1.\dfrac{1-\left(\dfrac{1}{3}\right)^9}{1-\left(\dfrac{1}{3}\right)}\approx1,5\)
b, Ta có \(\dfrac{\dfrac{1}{5}}{1}=\dfrac{1}{5};\dfrac{\dfrac{1}{25}}{\dfrac{1}{5}}=\dfrac{1}{5};...\)
-> Là cấp số nhân, q = 1/5
\(S_7=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}\approx1,25\)
Tính tổng các dãy số sau
a) S= \(1+0,1+\left(0,1\right)^2+\left(0,1\right)^3+...\)
b) S= \(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}+...\)
c) S= \(2+0,3+\left(0,3\right)^2+\left(0,3\right)^3+...\)
a. Dãy là tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=\dfrac{1}{10}\end{matrix}\right.\)
Do đó: \(S=\dfrac{u_1}{1-q}=\dfrac{1}{1-\dfrac{1}{10}}=\dfrac{10}{9}\)
b. Tương tự, tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-\dfrac{1}{3}\end{matrix}\right.\) bạn tự ráp công thức
c. \(S=2+S_1\) với \(S_1\) là cấp số nhân lùi vô hạn \(\left\{{}\begin{matrix}u_1=\dfrac{3}{10}\\q=\dfrac{3}{10}\end{matrix}\right.\)
Bài 1: Tính
a) \(\dfrac{7}{5}+3\dfrac{2}{5}-1\dfrac{1}{2}\)
b) 3 x 2\(\dfrac{4}{9}\)x\(\dfrac{3}{2}\)
c) \(\dfrac{5}{9}x\left(2\dfrac{1}{6}-1\dfrac{2}{3}\right)\)
`a/`
` 7/5 + 3 2/5 - 1 1/2 `
`= (7/5 + 17/5) - 3/2`
`= 24/5 - 3/2 `
`= 48/10 - 15/10 `
`= 33/10 `
`b/`
` 3 xx 2 4/9 xx 3/2 `
` = 3 xx 22/9 xx 3/2 `
` = 22/3 xx 3/2`
`= 11.`
`c/`
` 5/9 xx ( 2 1/6 - 1 2/3 ) `
`= 5/9 xx ( 13/6 -5/3 )`
`= 5/9 xx ( 13/6 - 10/6 ) `
`= 5/9 xx 3/6 `
`= 5/9 xx 1/2 `
`= 5/18`
Tính:
\(\dfrac{2}{5}+\dfrac{1}{5}\) \(\dfrac{2}{3}+\dfrac{5}{3}\) \(\dfrac{3}{8}+\dfrac{4}{8}\)
\(\dfrac{6}{9}+\dfrac{2}{9}\) \(\dfrac{12}{18}+\dfrac{7}{18}\) \(\dfrac{7}{4}+\dfrac{2}{4}\)
\(\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{2+1}{5}=\dfrac{3}{5}\)
\(\dfrac{2}{3}+\dfrac{5}{3}=\dfrac{2+5}{3}=\dfrac{7}{3}\)
\(\dfrac{3}{8}+\dfrac{4}{8}=\dfrac{3+4}{8}=\dfrac{7}{8}\)
\(\dfrac{6}{9}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}\)
\(\dfrac{12}{18}+\dfrac{7}{18}=\dfrac{12+7}{18}=\dfrac{19}{18}\)
\(\dfrac{7}{4}+\dfrac{2}{4}=\dfrac{7+2}{4}=\dfrac{9}{4}\)