Tính giá trị biểu thức:
B= \(\dfrac{\left(-2\right)^{24}.3^5-4^{12}.9^2}{8^8.3^5}+\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{301.303}\)
Tính giá trị của biểu thức sau:
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
1.tính giá trị biểu thức
a) \(\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
b) \(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).\left(1+\dfrac{1}{9.11}\right)\)
2. Chứng tỏ:
\(\dfrac{1}{201}+\dfrac{1}{202}+.........+\dfrac{1}{399}+\dfrac{1}{400}\)>\(\dfrac{1}{2}\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
Tính giá trị biểu thức:
\(e,\dfrac{18}{37}+\dfrac{8}{24}+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac{2}{3}\)
\(f,\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(g,\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}.\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(h,\dfrac{4}{9}.19\dfrac{1}{3}-\dfrac{4}{9}.39\dfrac{1}{3}\)
\(i,\left(-\dfrac{1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(j,125\%.\left(\dfrac{-1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2008^0\)
\(k,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(\dfrac{2n+5}{n+2}\)
\(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+L+\dfrac{1}{29}\)
\(A=\dfrac{7}{4}.\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}\)
\(A=\dfrac{17}{1.3}+\dfrac{17}{3.5}+\dfrac{17}{5.7}+...+\dfrac{17}{49.51}\)
\(\dfrac{2n+5}{3n+1}\)
\(\left(\left|x\right|+\dfrac{2}{5}\right):\dfrac{2}{5}=1\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}< 1\)
\(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
\(5.\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{10}\right)\le\dfrac{x}{20}\le-3\left(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{5}\right)\left(x\in Z\right)\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}< 1\)
trả hiểu yêu cầu đề bài là j cả
Tính R=\(\dfrac{\sqrt{\left(-\dfrac{2}{5}\right)^5.\left(-\dfrac{5}{8}\right)^3.5^2}}{\sqrt[3]{\left(-\dfrac{3}{4}\right)^3.\left(-\dfrac{5}{24}\right)^2.\left(-\dfrac{5}{3}\right)^4}}\)
\(R=\dfrac{\sqrt{\left(-\dfrac{2}{5}\cdot\dfrac{-5}{8}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-3^3}{4^3}\cdot\dfrac{5^2}{2^6\cdot3^2}\cdot\dfrac{5^4}{3^4}}}\)
\(=\dfrac{\sqrt{\left(\dfrac{1}{4}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-1}{3^3}\cdot\dfrac{25^3}{16^3}}}=\dfrac{5}{8}:\dfrac{-5}{3\cdot4}=\dfrac{5}{8}\cdot\dfrac{3\cdot4}{-5}=-\dfrac{3}{2}\)
Tính giá trị biểu thức.
a) \(\dfrac{4}{9}+\dfrac{2}{9}-\dfrac{5}{18}\) b) \(2-\dfrac{3}{5}+\dfrac{8}{15}\) c) \(\dfrac{9}{8}-\left(\dfrac{11}{8}-\dfrac{19}{32}\right)\)
a)
\(\dfrac{4}{9}+\dfrac{2}{9}-\dfrac{5}{18}\\ =\dfrac{6}{9}-\dfrac{5}{18}\\ =\dfrac{6\times2}{9\times2}-\dfrac{5}{18}\\ =\dfrac{12}{18}-\dfrac{5}{18}\\ =\dfrac{7}{18}\)
b)
\(2-\dfrac{3}{5}+\dfrac{8}{15}\\ =\dfrac{2\times15}{15}-\dfrac{3\times3}{5\times3}+\dfrac{8}{15}\\ =\dfrac{30-9+8}{15}\\ =\dfrac{29}{15}\)
c)
\(\dfrac{9}{8}-\left(\dfrac{11}{8}-\dfrac{19}{32}\right)\\ =\dfrac{9}{8}-\left(\dfrac{11\times4}{8\times4}-\dfrac{19}{32}\right)\\ =\dfrac{9}{8}-\left(\dfrac{44}{32}-\dfrac{19}{32}\right)\\ =\dfrac{9}{8}-\dfrac{25}{32}\\ =\dfrac{9\times4}{8\times4}-\dfrac{25}{32}\\ =\dfrac{36-25}{32}\\ =\dfrac{11}{32}\)
a) \(\dfrac{4}{9}+\dfrac{2}{9}-\dfrac{5}{18}=\dfrac{8}{18}+\dfrac{4}{18}-\dfrac{5}{18}=\dfrac{8+4-5}{18}=\dfrac{7}{18}\)
b) \(2-\dfrac{3}{5}+\dfrac{8}{15}=\dfrac{30}{15}-\dfrac{9}{15}+\dfrac{8}{15}=\dfrac{30-9+8}{15}=\dfrac{29}{15}\)
c) \(\dfrac{9}{8}-\left(\dfrac{11}{8}-\dfrac{19}{32}\right)=9-\left(\dfrac{44}{32}-\dfrac{19}{32}\right)=9-\dfrac{25}{32}=\dfrac{288}{32}-\dfrac{25}{32}=\dfrac{288-25}{32}=\dfrac{263}{32}\)
Bài 5: Tính nhanh tổng sau(nếu có):
M=\(\dfrac{3}{2}\)-\(\dfrac{5}{6}\)+\(\dfrac{7}{12}\)-\(\dfrac{9}{20}\)+\(\dfrac{11}{30}\)-\(\dfrac{13}{42}\)+\(\dfrac{15}{56}\)-\(\dfrac{17}{72}\) ; A=\(\dfrac{5}{1.3}\)+\(\dfrac{5}{3.5}\)+\(\dfrac{5}{5.7}\)+.....+\(\dfrac{5}{2019.2021}\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
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 - \(\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}\)
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\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
Tính giá trị của biểu thức:
\(A=\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(-2022\right)^0\)
\(B=0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2.\left(-\dfrac{1}{9}\right)\)
\(C=2\dfrac{6}{7}.\left[\left(\dfrac{-7}{5}-\dfrac{3}{2}:\dfrac{-5}{-4}\right)+\left(\dfrac{3}{2}\right)^2\right]\)
\(D=\dfrac{2}{7}+\dfrac{5}{7}.\left(\dfrac{3}{5}-0,25\right).\left(-2\right)^2+35\%\)
\(E=1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):1\dfrac{2}{5}\)
\(F=\dfrac{\dfrac{5}{3}-\dfrac{5}{7}+\dfrac{5}{9}}{\dfrac{10}{3}-\dfrac{10}{7}+\dfrac{10}{9}}\)