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1/ Bg
\(\frac{21^4}{27.\left(-343\right)}\)= \(\frac{\left(3.7\right)^4}{3^3.\left(-7\right)^3}\)
= \(\frac{3^4.7^4}{3^3.\left(-7\right)^3}\)
= \(\frac{3.\left(-7\right)}{1.1}\)
= 3.(-7)
= -21
2/ Bg
Ta có: \(\frac{a}{4}=\frac{b}{5}=\frac{c}{2}\)và a + b - c = 21 (a, b, c thuộc Z)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{2}=\frac{a+b-c}{4+5-2}=\frac{21}{7}\)= 3
=> a = 3.4 = 12
=> b = 3.5 = 15
=> c = 3.2 = 6
Vậy a = 12, b = 15 và c = 6
a) \(625^4:25^7\)
\(=\left[25^2\right]^4:25^7\)
\(=25^8:25^7\)
\(=25\)
b)\(\left(100^5-89^5\right).\left(6^8-8^6\right).\left(8^2-4^3\right)\)
\(=\left(100^5-89^5\right).\left(6^8-8^6\right).\left[\left(2^3\right)^2-\left(2^2\right)^3\right]\)
\(=\left(100^5-89^5\right).\left(6^8-8^6\right).\left[2^6-2^6\right]\)
\(=\left(100^5-89^5\right).\left(6^8-8^6\right).0\)
\(=0\)
a) A = \(\frac{101}{19}.\) \(\frac{61}{218}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
= \(\frac{101}{218}.\frac{61}{19}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.\left(\frac{61}{19}-\frac{42}{19}\right)+\frac{117}{218}\)
=\(\frac{101}{218}.\frac{19}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.1+\frac{117}{218}\)
=\(\frac{101}{218}+\frac{117}{218}\)
=\(\frac{218}{218}\)\(=1\)
b) B = \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\)\(.\left(\frac{1}{20}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).0\)
= \(0\)
a) \(\frac{1}{3}-\frac{-1}{6}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)
b) \(2\frac{1}{3}+4\frac{1}{5}=\frac{7}{3}+\frac{21}{5}=\frac{98}{15}\)
c) \(\frac{4}{9}-\frac{13}{3}-\frac{4}{9}-\frac{10}{3}=\left(\frac{4}{9}-\frac{4}{9}\right)-\left(\frac{13}{3}+\frac{10}{3}\right)\)
\(=0-\frac{23}{3}=\frac{-23}{3}\)
d) \(4-\left(2-\frac{5}{2}\right)+0,5=4-2+\frac{5}{2}+\frac{1}{2}=2+3=5\)
=\(\frac{3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{49\cdot51}{50\cdot50}\)
=\(\frac{\left(2\cdot3\cdot...\cdot49\right)\cdot\left(3\cdot4\cdot...\cdot51\right)}{\left(2\cdot3\cdot4\cdot...\cdot50\right)\left(2\cdot3\cdot4\cdot...\cdot50\right)}\)
=\(\frac{51}{50\cdot2}=\frac{51}{100}\)
\(a)\frac{x}{8}=\frac{-30}{y}=\frac{-48}{32}\)
Rút gọn : \(\frac{-48}{32}=\frac{(-48):16}{32:16}=\frac{-3}{2}\)
* Ta có : \(\frac{x}{8}=\frac{-3}{2}\)
\(\Rightarrow x\cdot2=-3\cdot8\)
\(\Rightarrow x=\frac{-3\cdot8}{2}=-12\)
* Ta có : \(\frac{-30}{y}=\frac{-3}{2}\)
\(\Rightarrow-30\cdot2=-3\cdot y\)
\(\Rightarrow y=\frac{-30\cdot2}{-3}=20\)
Mấy bài kia làm tương tự
Đáp án cần chọn là: D
Ta có (56.35+56.18):53=56.(35+18):53=56.53:53=56.1=56