
\(\left(\frac{7}{2}\right)^{-2}\).\...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a: \(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\) b: \(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{1}{2}\cdot\dfrac{6}{5}-17=\dfrac{3}{5}-17=-\dfrac{82}{5}\) c: \(=-\left(\dfrac{1}{3}\right)^{50}\cdot3^{50}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\) e: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\) Bài 1: a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\) \(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\) \(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\) \(=\frac{-2}{2}=-1\) b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\) \(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\) \(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\) \(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\) c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\) \(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\) \(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\) d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\) \(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\) \(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\) \(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\) e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\) \(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\) \(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\) \(=\frac{11}{12}\) f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\) \(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\) \(=\frac{1}{5}\) Bài 1 \(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\) \(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\) \(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\) \(=\frac{9}{25}+\frac{8}{9}-1\) \(=\frac{56}{225}\) \(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\) \(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\) \(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\) \(=1:\frac{4}{3}=\frac{3}{4}\) Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v \(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\) \(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\) \(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\) \(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\) \(=-\frac{1}{2}\) a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\) = \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\) = \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\) =\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\) =\(\frac{1}{2}-1-1\) =\(\frac{-3}{2}\) b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\) = \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\) = \(-12:\left\{\frac{-1}{12}\right\}^2\) = \(-12:\frac{1}{144}\) = \(-12.144\) = -1728 c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\) = \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\) = \(\frac{7}{23}.\left(\frac{-23}{6}\right)\) = \(\frac{-7}{6}\) d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\) = \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\) = \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\) = \(10.\frac{7}{5}\) = 14 e) (1+23−14).(0,8−34)2 = (1+23−14).(\(\frac{4}{5}\)−34)2 = \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\) = \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\) = \(\frac{17}{20}.\frac{1}{400}\) = \(\frac{17}{8000}\) Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\) =\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\) \(=\frac{5}{2}.\frac{6}{5}-17\) \(=3-17=-14\) a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\) \(=1-\frac{1}{100}< 1\) \(\Rightarrow A< 1\) b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\) \(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\) \(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\) \(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\) \(\Rightarrow2B< 1\) \(\Rightarrow B< \frac{1}{2}\)
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