Tính:
\(\dfrac{8^3.5^4}{4^5.125}\)
Tính giá trị biểu thức sau (hợp lí nếu có thể)
a) \(\dfrac{\left(-3\right)^7.2^8}{6^7}\) b) \(\dfrac{5^3.3^5}{5^3.0,5+125.2,5}\)
c) \(\dfrac{5.7^4+7^3.25}{7^5.125-7^3.50}\)
-3^7.2^8/2^.3^7
=-3.2
=-6
5^3.3^5/5^3(0,5+2,5)
=5^3.3^5/5^3.3\
3^4
=81
5.7^4+7^3.25/7^5.125-7^3.50
=5.7^3(7+5
5.7^4+7^3.25/7^5.125-7^3.50
=5.7^4+7^3.5^2/7^5.5^3-7^3.11.5
=5.7^3(1.7+1.5)/7^3.5(7^2.25-11)
12/1250
A=\(\dfrac{4}{3.5}-\dfrac{6}{5.7}+\dfrac{8}{7.9}-\dfrac{10}{9.11}+\dfrac{12}{11.13}-...-\dfrac{100}{99.100}\)
Tính giá trị của A
Tính R=\(\dfrac{\sqrt{\left(-\dfrac{2}{5}\right)^5.\left(-\dfrac{5}{8}\right)^3.5^2}}{\sqrt[3]{\left(-\dfrac{3}{4}\right)^3.\left(-\dfrac{5}{24}\right)^2.\left(-\dfrac{5}{3}\right)^4}}\)
\(R=\dfrac{\sqrt{\left(-\dfrac{2}{5}\cdot\dfrac{-5}{8}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-3^3}{4^3}\cdot\dfrac{5^2}{2^6\cdot3^2}\cdot\dfrac{5^4}{3^4}}}\)
\(=\dfrac{\sqrt{\left(\dfrac{1}{4}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-1}{3^3}\cdot\dfrac{25^3}{16^3}}}=\dfrac{5}{8}:\dfrac{-5}{3\cdot4}=\dfrac{5}{8}\cdot\dfrac{3\cdot4}{-5}=-\dfrac{3}{2}\)
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
tính hợp lý:
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)
`4/1.3+4/3.5+4/5.7+...+4/99.101`
`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`
`=2(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`
`=2(1-1/101)`
`=2. 100/101`
`=200/101`
Tính giá trị biểu thức:
B= \(\dfrac{\left(-2\right)^{24}.3^5-4^{12}.9^2}{8^8.3^5}+\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{301.303}\)
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101
\(M=\left(\dfrac{4}{1.3}-\dfrac{8}{3.5}+\dfrac{12}{5.7}-...+\dfrac{4028}{2013.2015}\right).\dfrac{2015}{2016}\)
Chứng minh M là số tự nhiên.
Trong dấu ngoặc đơn có số các số hạng là
Đặt tổng các số hạng trong ngoặc đơn là A
\(\dfrac{2013-1}{2}+1=1007\) số hạng
\(A=\dfrac{3+1}{1.3}-\dfrac{5+3}{3.5}+\dfrac{7+5}{5.7}-...+\dfrac{2015+2013}{2013.2015}=\)
\(=1+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}-...+\dfrac{1}{2013}+\dfrac{1}{2015}=1+\dfrac{1}{2015}=\dfrac{2016}{2015}\)
\(\Rightarrow M=A.\dfrac{2015}{2016}=\dfrac{2016}{2015}.\dfrac{2015}{2016}=1\) là số tự nhiên
giúp mik vs
\(\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.7}+.....+\dfrac{4^2}{45.47}.\dfrac{1-3-5-..-49}{8}\) bài này tính nha
\(\dfrac{4}{3.5}\) + \(\dfrac{8}{5.9}\) + \(\dfrac{12}{9.15}\)+.......+
\(\dfrac{32}{x(x+16)}\) = \(\dfrac{16}{15}\)
\(\dfrac{32}{x(x+16)}\) = \(\dfrac{16}{15}\)
\(\dfrac{HeLp}{me}\) \(\dfrac{nguyễn}{đức}{Trí}\)
\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)
\(X+16=-5\)
\(X=-21\)
Tính hợp lý:
\(\dfrac{4^2.25^2+32.125}{2^3.5^2}\)
\(\dfrac{4^2.25^2+32.125}{2^3.5^2}=\dfrac{2^4.5^4+2^5.5^3}{2^3.5^2}=\dfrac{2^3.5^2\left(2.5^2+2^2.5\right)}{2^3.5^2}=2.5^2+2^2.5=50+20=70\)