c,\(2021,2345\cdot2020,1234+2021,2345\cdot\left(-2020,1234\right)\)
d,\(4,75+\left(\dfrac{-1}{2}\right)^3+0,5^2-3\cdot\dfrac{-3}{8}\)
Những câu hỏi liên quan
Tính một cách hợp lí:
\(\begin{array}{l}a)A = 32,125 - (6,325 + 12,125) - (37 + 13,675)\\b)B = 4,75 + {\left( {\frac{{ - 1}}{2}} \right)^3} + 0,{5^2} - 3.\frac{{ - 3}}{8}\\c)C = 2021,2345.2020,1234 + 2021,2345.( - 2020,1234)\end{array}\)
\(\begin{array}{l}a)A = 32,125 - (6,325 + 12,125) - (37 + 13,675)\\ = 32,125 - 6,325 - 12,125 - 37 - 13,675\\ = (32,125 - 12,125) + ( - 6,325 - 13,675) - 37\\ = 20 + ( - 20) - 37\\ = - 37\\b)B = 4,75 + {\left( {\frac{{ - 1}}{2}} \right)^3} + 0,{5^2} - 3.\frac{{ - 3}}{8}\\ = 4,75 + \frac{{ - 1}}{8} + 0,25 + \frac{9}{8}\\ = (4,75 + 0,25) + \left( {\frac{{ - 1}}{8} + \frac{9}{8}} \right)\\ = 5 + \frac{8}{8}\\ = 5 + 1\\ = 6\\c)C = 2021,2345.2020,1234 + 2021,2345.( - 2020,1234)\\ = 2021,2345.[2020,1234 + ( - 2020,1234)]\\ = 2021,2345.0\\ = 0\end{array}\)
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Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
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1/Sleft(1+dfrac{1}{2}right)cdotleft(1+dfrac{1}{3}right)cdotleft(1+dfrac{1}{4}right)cdot...cdotleft(1+dfrac{1}{100}right)
2/Bleft(1-dfrac{1}{2}right)cdotleft(1-dfrac{1}{3}right)cdotleft(1-dfrac{1}{4}right)cdot...cdotleft(1-dfrac{1}{2007}right)
3/Cdfrac{2^2}{1cdot3}cdotdfrac{3^2}{2cdot4}cdotdfrac{4^2}{3cdot5}cdot...cdotdfrac{100^2}{99cdot101}
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1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
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Tính:
\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
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Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $mathrm{A}left(dfrac{2}{7} cdot dfrac{1}{4}-dfrac{1}{3} cdot dfrac{2}{7}right):left(dfrac{2}{7} cdot dfrac{3}{9}-dfrac{2}{7} cdot dfrac{2}{5}right)$;
b) $mathrm{B}dfrac{left(dfrac{1}{5}-dfrac{2}{7}right) cdot dfrac{3}{4}-dfrac{3}{4} cdotleft(dfrac{1}{3}-dfrac{2}{7}right)}{dfrac{1}{5} cdot dfrac{2}{7}-dfrac{1}{3} cdotleft(dfrac{2}{7}+dfrac{3}{9}right)+dfrac{3}{9} cdot dfrac{1}{5}} .$
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Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$;
b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}} .$
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Tính giá trị của các biểu thức sau 1) A1+2+2^2+...+2^{2015} 2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right) 3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89} 4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3} 5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}...
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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 tích và viết kết quả ở dạng phân số tối giản:
a) \(\dfrac{-5}{9}\cdot\dfrac{12}{35};\)
b) \(\left(-\dfrac{5}{8}\right)\cdot\dfrac{-6}{55};\)
c) \(\left(-7\right)\cdot\dfrac{2}{5};\)
d) \(\dfrac{-3}{8}\cdot\left(-6\right).\)
\(a.\)
\(-\dfrac{5}{9}\cdot\dfrac{12}{35}=\dfrac{\left(-5\right)\cdot12}{9\cdot35}=\dfrac{-60}{315}=-\dfrac{4}{21}\)
\(b.\)
\(\left(-\dfrac{5}{8}\right)\cdot-\dfrac{6}{55}=\dfrac{\left(-5\right)\cdot\left(-6\right)}{8\cdot55}=\dfrac{30}{440}=\dfrac{3}{44}\)
\(c.\)
\(\left(-7\right)\cdot\dfrac{2}{5}=-\dfrac{14}{5}\)
\(d.\)
\(-\dfrac{3}{8}\cdot\left(-6\right)=\dfrac{-3\cdot\left(-6\right)}{8}=\dfrac{18}{8}=\dfrac{9}{4}\)
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a) \(\dfrac{-5}{9}.\dfrac{12}{35}=\dfrac{-5.12}{9.35}=\dfrac{-4}{21}\)
b) \(\dfrac{-5}{8}.\dfrac{-6}{55}=\dfrac{-5.-6}{8.55}=\dfrac{3}{44}\)
c)\(-7.\dfrac{2}{5}=\dfrac{-7.2}{5}=\dfrac{-14}{5}\)
d) \(\dfrac{-3}{8}.-6=\dfrac{-3.-6}{8}=\dfrac{9}{4}\)
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\(\dfrac{-5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\)
\(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\)
\(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
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\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
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\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{15}.3+\left(\dfrac{32}{60}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}:\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}.\dfrac{24}{47}\\ =\dfrac{7}{5}+\dfrac{222}{235}\\ =\dfrac{329}{235}+\dfrac{222}{235}\\ =\dfrac{551}{235}\)
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2 tháng 12 2023 lúc 20:18
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0.5\cdot\left(-2\dfrac{3}{5}\right)\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)
\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)
\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)
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