Cho biểu thức : Tính nhanh :
A = \(\dfrac{1+5+5^2+5^3+..........+5^9}{1+5+5^2+5^3+...........+5^8}\)
B = \(\dfrac{1+3+3^2+3^3+..........+3^9}{1+3+3^2+3^3+..............+3^8}\)
Cho biểu thức :
A = \(\dfrac{1+5+5^2+..........+5^9}{1+5+5^2+............+5^8}\)
B = \(\dfrac{1+3+3^2+............+3^9}{1+3+3^2+............+3^8}\)
Giúp mình nhanh với .
Theo bài ra, ta có:
+) A = \(\dfrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)
= \(\dfrac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}\)+ \(\dfrac{5^9}{1+5+5^2+...+5^8}\)
= 1 + \(\dfrac{1}{\dfrac{1+5+5^2+...+5^8}{5^9}}\)
+) B = \(\dfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
= \(\dfrac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}\)+ \(\dfrac{3^9}{1+3+3^2+...+3^8}\)
= 1 + \(\dfrac{1}{\dfrac{1+3+3^2+...+3^8}{3^9}}\)
Nhận xét:
+) \(\dfrac{1+5+5^2+...+5^8}{5^9}\) = \(\dfrac{1}{5^9}\) + \(\dfrac{1}{5^8}\) + ... + \(\dfrac{1}{5^{ }}\)
+) \(\dfrac{1+3+3^2+...+3^8}{3^9}\) = \(\dfrac{1}{3^9}\) + \(\dfrac{1}{3^8}\) + ... + \(\dfrac{1}{3}\)
Có: \(\dfrac{1}{5^9}\) < \(\dfrac{1}{3^9}\) ; \(\dfrac{1}{5^8}\) < \(\dfrac{1}{3^8}\) ; ... ; \(\dfrac{1}{5^{ }}\) < \(\dfrac{1}{3}\)
⇒ \(\dfrac{1+5+5^2+...+5^8}{5^9}\) < \(\dfrac{1+3+3^2+...+3^8}{3^9}\)
⇒ \(\dfrac{1}{\dfrac{1+5+5^2+...+5^8}{5^9}}\) > \(\dfrac{1}{\dfrac{1+3+3^2+...+3^8}{3^9}}\)
⇒ A > B
Vậy A > B.
Thực hiện phép tính ( tính nhanh nếu có thể )
a, -5/21 + -2/21 + 8/24
b, 4/11 . -2/7 + 4/11 . -4/7 + 4/11 . -1/7
c, \(10\dfrac{5}{9}\) - ( \(3\dfrac{5}{7}\) + \(4\dfrac{5}{9}\) )
d, 1/3 + \(1\dfrac{3}{4}\) - ( \(1\dfrac{3}{4}\) - 80% )
e, \(5\dfrac{3}{5}\) + \(7\dfrac{21}{48}\) : 10/7 - \(5\dfrac{21}{48}\) : 10/7
f, -5/7 . 2/11 - 5/11 . 9/7 + \(2\dfrac{5}{7}\)
g, -3/13 . 6/8 + 7/13 . -3/8 + \(1\dfrac{3}{8}\)
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d:
12+3/7-11+3/7 tính bằng cách thuận tiện nhé
Tính:
\(\dfrac{1}{3}+\dfrac{2}{9}\) \(\dfrac{1}{2}+\dfrac{3}{8}\) \(\dfrac{5}{12}+\dfrac{2}{3}\)
\(\dfrac{5}{16}+\dfrac{3}{8}\) \(\dfrac{4}{15}+\dfrac{3}{5}\) \(\dfrac{8}{63}+\dfrac{7}{10}\)
\(\dfrac{1}{3}+\dfrac{2}{9}=\dfrac{3}{3\times3}+\dfrac{2}{9}=\dfrac{3}{9}+\dfrac{2}{9}=\dfrac{5}{9}\)
\(\dfrac{1}{2}+\dfrac{3}{8}=\dfrac{4}{2\times4}+\dfrac{3}{8}=\dfrac{4}{8}+\dfrac{3}{8}=\dfrac{7}{8}\)
\(\dfrac{5}{12}+\dfrac{2}{3}=\dfrac{5}{12}+\dfrac{2\times4}{3\times4}=\dfrac{5}{12}+\dfrac{8}{12}=\dfrac{13}{12}\)
\(\dfrac{5}{16}+\dfrac{3}{8}=\dfrac{5}{16}+\dfrac{3\times2}{8\times2}=\dfrac{5}{16}+\dfrac{6}{16}=\dfrac{11}{16}\)
\(\dfrac{4}{15}+\dfrac{3}{5}=\dfrac{4}{15}+\dfrac{3\times3}{5\times3}=\dfrac{4}{15}+\dfrac{9}{15}=\dfrac{13}{15}\)
\(\dfrac{8}{63}+\dfrac{7}{10}=\dfrac{8\times10}{63\times10}+\dfrac{7\times63}{10\times63}=\dfrac{80}{630}+\dfrac{441}{630}=\dfrac{521}{630}\)
Tính (theo mẫu):
a)
\(5+\dfrac{3}{2}\) \(\dfrac{3}{4}+2\) \(\dfrac{8}{9}+3\)
b)
\(1-\dfrac{1}{2}\) \(5-\dfrac{7}{3}\) \(\dfrac{11}{2}-3\)
a: \(5+\dfrac{3}{2}=\dfrac{10}{2}+\dfrac{3}{2}=\dfrac{10+3}{2}=\dfrac{13}{2}\)
\(\dfrac{3}{4}+2=\dfrac{3}{4}+\dfrac{8}{4}=\dfrac{3+8}{4}=\dfrac{11}{4}\)
\(\dfrac{8}{9}+3=\dfrac{8}{9}+\dfrac{27}{9}=\dfrac{8+27}{9}=\dfrac{35}{9}\)
b: \(1-\dfrac{1}{2}=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{2-1}{2}=\dfrac{1}{2}\)
\(5-\dfrac{7}{3}=\dfrac{15}{3}-\dfrac{7}{3}=\dfrac{15-7}{3}=\dfrac{8}{3}\)
\(\dfrac{11}{2}-3=\dfrac{11}{2}-\dfrac{6}{2}=\dfrac{11-6}{2}=\dfrac{5}{2}\)
rút gọn các biểu thức sau:
a \(\sqrt[3]{8\sqrt{5}-16}.\sqrt[3]{8\sqrt{5}+16}\)
b \(\sqrt[3]{7-5\sqrt{2}}-\sqrt[6]{8}\)
c \(\sqrt[3]{4}.\sqrt[3]{1-\sqrt{3}}.\sqrt[6]{4+2\sqrt{3}}\)
d \(\dfrac{2}{\sqrt[3]{3}-1}-\dfrac{4}{\sqrt[3]{9}-\sqrt[3]{3}+1}\)
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
Tính giá trị của biểu thức sau:
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
Tính giá trị của các biểu thức sau :
a)\(\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)\)+(0,4 - 5) - \(\left(4\dfrac{1}{4}-1\right)\)
b)\(\dfrac{2}{3}\) - \(\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
c)\(\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right)\):\(\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
d)3 - \(\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}\)
giúp mình nhé trả lời mình cho tick cảm ơn các bạn !
\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
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
Tính giá trị của biểu thức:
\(A=\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(-2022\right)^0\)
\(B=0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2.\left(-\dfrac{1}{9}\right)\)
\(C=2\dfrac{6}{7}.\left[\left(\dfrac{-7}{5}-\dfrac{3}{2}:\dfrac{-5}{-4}\right)+\left(\dfrac{3}{2}\right)^2\right]\)
\(D=\dfrac{2}{7}+\dfrac{5}{7}.\left(\dfrac{3}{5}-0,25\right).\left(-2\right)^2+35\%\)
\(E=1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):1\dfrac{2}{5}\)
\(F=\dfrac{\dfrac{5}{3}-\dfrac{5}{7}+\dfrac{5}{9}}{\dfrac{10}{3}-\dfrac{10}{7}+\dfrac{10}{9}}\)