(1,5 điểm) Tính
a) $0,2+1\dfrac{3}{7}-\dfrac{6}{5}$;
b) $\left( \dfrac{4}{5}-1 \right) \, : \, \dfrac{3}{5}-\dfrac{2}{3}. 0,5$.
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1)
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{6}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
\(=\dfrac{3}{-5}+\dfrac{3}{5}\)
\(=-\dfrac{3}{5}+\dfrac{3}{5}\)
\(=0\)
a) x + 2/5 = -4/3
x = -4/3 - 2/5
x = -26/15
b) -5/6 + 1/3 x = (-1/2)²
-5/6 + 1/3 x = 1/4
1/3 x = 1/4 + 5/6
1/3 x = 13/12
x = 13/12 : 1/3
x = 13/4
c) 7/12 - (x + 7/6) . 6/5 = (-1/2)³
7/12 - (x + 7/6) . 6/5 = -1/8
(x + 7/6) . 6/5 = 7/12 + 1/8
(x + 7/6) . 6/5 = 17/24
x + 7/6 = 17/24 : 6/5
x + 7/6 = 85/144
x = 85/144 - 7/6
x = -83/144
\(a,x+\dfrac{2}{5}=-\dfrac{4}{3}\\ \Rightarrow x=-\dfrac{26}{15}\\ b,-\dfrac{5}{6}+\dfrac{1}{3}x=\left(-\dfrac{1}{2}\right)^2\\ \Rightarrow-\dfrac{5}{6}+\dfrac{1}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{3}x=\dfrac{13}{12}\\ \Rightarrow x=\dfrac{13}{4}\\ c,\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\left(-\dfrac{1}{2}\right)^3\\ \Rightarrow\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=-\dfrac{1}{8}\\ \Rightarrow\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\dfrac{17}{24}\\ \Rightarrow x+\dfrac{7}{6}=\dfrac{85}{144}\\ \Rightarrow x=-\dfrac{83}{144}.\)
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0
A = 0*
*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)
\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
a. \(\left(\dfrac{-2}{7}+\dfrac{4}{7}\right)+\dfrac{1}{7}=\dfrac{2}{7}+\dfrac{1}{7}=\dfrac{3}{7}\)
b. \(\left(\dfrac{-8}{19}+\dfrac{27}{19}\right)+\left(\dfrac{-4}{21}+\dfrac{-17}{21}\right)+\dfrac{-12}{16}=\dfrac{19}{19}+\dfrac{-21}{21}+\dfrac{-12}{16}\)
\(=1+\left(-1\right)+\dfrac{-12}{16}=\dfrac{-12}{16}\)
c. \(\dfrac{-3}{7}+\dfrac{-11}{13}+\dfrac{-2}{13}+\dfrac{3}{7}+1=\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{-11}{13}+\dfrac{-2}{13}\right)+1=0-1+1=0\)
-2/7 + 2/7 : 3/5
= -2/7 + 2/7. 3/5
= -2/7 + 35/6
= -12/42 + 245/42
= -233/42
= (-4/21 - 17/21) + (-8/19 + 27/19)
= - 13/21 + -19/19
= (-13/21)+ (-1)
= (-13/21) + (-1/1)
= (-13/21) + (-13/21)
= - 26/21
=6/5. (3/13 - 16/13)
= 6/5. (-13/13)
=6/5. (-1)
=6/5. (-1/1)
=6/5. (-6/5)
=(-36/25)
a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{17}{35}\) + 0,4
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{3}{35}\)
\(x\) = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)
\(x\) = - \(\dfrac{1}{5}\)
b, 0,2.(\(x\) - 3) +2,4 = 10
0,2.(\(x\) - 3) = 10 - 2,4
0,2.(\(x\) - 3) = 7,6
\(x\) - 3 = 7,6:0,2
\(x\) - 3 = 38
\(x\) = 38 + 3
\(x\) = 41
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
\(A=\left(-1,5\right)^2\cdot2\dfrac{2}{3}-\dfrac{1}{6}+\left(\dfrac{4}{7}-\dfrac{2}{5}\right):1\dfrac{1}{35}\)
\(=\left(-\dfrac{3}{2}\right)^2\cdot\dfrac{8}{3}-\dfrac{1}{6}+\left(\dfrac{20}{35}-\dfrac{14}{35}\right):\dfrac{36}{35}\\ =\dfrac{9}{4}\cdot\dfrac{8}{3}-\dfrac{1}{6}+\dfrac{6}{35}\cdot\dfrac{35}{36}\\ =6-\dfrac{1}{6}+\dfrac{1}{6}\\ =6\)
\(a)0,2-1\dfrac{3}{7}-\dfrac{6}{5}\)
\(=\dfrac{1}{5}-\dfrac{10}{7}-\dfrac{6}{5}\)
\(=(\dfrac{1}{5}-\dfrac{6}{5}\)-\dfrac{10}{7}\)
\(=-1-\dfrac{10}{7}\)
\(=\dfrac{-17}{7}\)
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\(b)(\dfrac{4}{5}-1):\dfrac{3}{5}-\dfrac{2}{3}.0,5\)
\(=\dfrac{-1}{5}.\dfrac{5}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{-1}{3}-\dfrac{1}{3}=\dfrac{-2}{3}\)