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Bài 1:
a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)
\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)
\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)
\(=-10\cdot1.7=-17\)
b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)
\(=-\dfrac{539}{60}\)
c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
\(=10\)
\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}-\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{115}{-161}=-\dfrac{115}{161}\)
a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)
= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)
= \(2,75.\left(-3,2\right)\)
= \(-8,8\)
b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)
= \(\dfrac{3}{7}-\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)
= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)
= \(-\dfrac{23}{20}\)
d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)
= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)
=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)
= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)
= \(-\dfrac{23}{40}\)
e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)
= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)
= \(-\dfrac{5501}{225}\)
a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)
\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)
\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)
\(=\dfrac{-54}{5}\)
b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)
\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)
\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)
\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
Giải:
a) \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
Vậy giá trị của biểu thức trên là \(\dfrac{1}{100}\).
b) \(\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{4}{11}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{44}{121}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\dfrac{0}{121}\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
c) \(-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}\right)-\dfrac{2}{5}\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left[\left(\dfrac{15}{17}-\dfrac{9}{15}\right)+\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\right]\)
\(=-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}+\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left(1-1\right)\)
\(=-\dfrac{2}{5}.0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
Chúc bạn học tốt!!!
\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^3}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}....\dfrac{99.99}{99.100}\)
\(=\dfrac{1.1.2.2.3.3.....99.99}{1.2.2.3.3.4....99.100}\)
\(=\dfrac{1.2.3...99}{1.2.3....99}.\dfrac{1.2.3....99}{2.3.4....100}=1.\dfrac{1}{100}=\dfrac{1}{100}\)
= 11/4. ( -4/10) - 8/3 . 275/100 + ( - 12/10) : 4/11
= -9/10 - 21/3+ ( -27/10)
= [-9/10+( -27/10)] - 21/3
= -12/5 - 21/3
= -43/5
2$\dfrac{}{4}$ . (-0,4) - 1$\dfrac{3}{5}$ . 2,75 + (-1,2) : $\dfrac{4}{11}$
= $\dfrac{11}{4}$ . (-0,4) - $\dfrac{8}{5}$ . 2,75 + (-1,2) : $\dfrac{4}{11}$
= $\dfrac{-11}{10}$ - $\dfrac{22}{5}$ + $\dfrac{-33}{10}$
= ($\dfrac{-11}{10}$ - $\dfrac{22x2}{5x2}$ + $\dfrac{-33}{10}$
= ($\dfrac{-11}{10}$ - $\dfrac{44}{10}$ + $\dfrac{-33}{10}$
= $\dfrac{(-11)-44+(-33)}{10}$
= $\dfrac{-44}{5}$.