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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) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)
= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)
= \(-\dfrac{5501}{225}\)
e) \(\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}.\dfrac{5}{7}+2\dfrac{3}{5}\)
= \(\dfrac{-3}{5}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}.1+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}+\dfrac{13}{5}\)
= 2
a) -1,6: (1+2/3)-76/25 =
-1,6:5/3-76/25=
-24/25-76/25=-100/25=-4
b) -3+1/5 . 5/64 + (0,8-(2+4/15)) : 11/3=
-16/5.5/64+(12/15-34/15):11/3=
-1/4+(-22/15):11/3=
-1/4+(-2/5)=
3/20
tick cho mình nha các bạn
\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)
\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)
\(=\dfrac{231}{54}:\dfrac{7}{12}\)
\(=\dfrac{198}{27}\)
\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)
\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)
\(=\dfrac{0,8.0,64}{0,6}\)
\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
\(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}\)
\(=\dfrac{3}{4}+\dfrac{-2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)
1) \(-\dfrac{5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =-\dfrac{5}{9}+\dfrac{14}{9}\cdot\dfrac{7}{20}\cdot\dfrac{1}{49}\\ =-\dfrac{5}{9}+\dfrac{1}{90}=\dfrac{-49}{90}\)
2) \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right)\cdot\dfrac{24}{47}-\dfrac{51}{13}\cdot\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =-\dfrac{4}{13}\)
3) \(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{2}{5}\\ =1\)
Tìm x : 1) \(60\%x+0,4x+x:3=2\\ \Leftrightarrow\dfrac{3}{5}x+\dfrac{2}{5}x+\dfrac{1}{3}x=2\\ \Leftrightarrow\dfrac{4}{3}x=2\\ \Leftrightarrow x=\dfrac{3}{2}\)
Nốt nè bn
\(-2x-\dfrac{-3}{5}:\left(0,5\right)^2=-1\dfrac{1}{4}\\ \Leftrightarrow-2x+\dfrac{3}{5}:\dfrac{1}{4}=-\dfrac{5}{4}\\ \Leftrightarrow-2x+\dfrac{12}{5}=-\dfrac{5}{4}\\ \Leftrightarrow-2x=-\dfrac{73}{20}\\ x=-\dfrac{73}{40}\)
\(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{2}{3}-\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{31}{60}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
1) \(25\%-1\dfrac{1}{2}+0,5\cdot\dfrac{12}{5}\\ =\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}\cdot\dfrac{12}{5}\\ =-\dfrac{5}{4}+\dfrac{6}{5}\\ =-\dfrac{1}{20}\)
2) \(\left(-3,2\right)\cdot\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{1}{2}\)
\(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{7}{2}\\ =\dfrac{3}{4}-\dfrac{22}{15}\cdot\dfrac{2}{7}\\ =\dfrac{3}{4}-\dfrac{44}{105}\\ =\dfrac{139}{420}\)