Tính :
\(C=\left(0,5\right)^{-4}-625^{0,25}-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}+19\left(-3\right)^{-3}\)
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GV
Giáo viên Toán
Giáo viên
26 tháng 4 2017
a) \(\left(\dfrac{1}{16}\right)^{-\dfrac{3}{4}}+810000^{0.25}-\left(7\dfrac{19}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{4.\left(-\dfrac{3}{4}\right)}+\left(30\right)^{4.0,25}-\left(\dfrac{243}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{-3}+30-\left(\dfrac{3}{2}\right)^{5.\dfrac{1}{5}}\)
\(=2^3+30-\dfrac{3}{2}\)
\(=36,5\)
GV
Giáo viên Toán
Giáo viên
26 tháng 4 2017
b) \(=\left(0,1\right)^{3.\left(-\dfrac{1}{3}\right)}-2^{-2}.2^{6.\dfrac{2}{3}}-\left[\left(2\right)^3\right]^{-\dfrac{4}{3}}\)
\(=0,1^{-1}-2^2-2^{-4}\)
\(=10-4-\dfrac{1}{16}\)
\(=\dfrac{95}{16}\)
20 tháng 1 2020
a) 4x - 3 = 4- 3x
<=> 4x + 3x = 4 + 3
<=> 7x = 7
<=> x = 1
b) 3 + (x - 5) = 2 ( 3x - 2)
<=> 3 + x - 5 = 6x - 4
,<=> x- 2 = 6x - 4
<=> 4 - 2 = 6x - x
<=> 2 = 5x
,<=> 5x = 2
<=> x = \(\frac{2}{5}\)
c) 2( x - \(\frac{1}{4}\)) - 4 = -6 ( -\(\frac{1}{3}\)+ 0,5) + 2
<=> 2x -\(\frac{1}{2}\)- 4 = 2 - 3 + 2
<=> 2x- \(\frac{9}{2}\)= 1
,<=> 2x = 1 + \(\frac{9}{2}\)= \(\frac{11}{2}\)
<=> x = \(\frac{11}{4}\)
d) 2 ( x - 0,5) + 3 = 0,25 ( 4x - 1)
<=> 2x - 1 + 3 = x - 0,25
<=> 2x + 2 = x - 0,25
<=> 2x - x = -2 - 0,25
<=> x = -2
1 tháng 4 2018
\(A=\left(0,25\right)^{-1}.\left(\frac{1}{4}\right)^{-2}.\left(\frac{4}{3}\right)^2.\left(\frac{5}{4}\right)^{-1}.\left(\frac{2}{3}\right)^{-3}\)
\(\Rightarrow A=4^1.4^2.\frac{16}{9}.\frac{4}{5}\frac{27}{8}\)
\(\Rightarrow A=\frac{64}{1}.\frac{16}{9}.\frac{4}{5}.\frac{27}{8}\)
\(\Rightarrow A=\frac{1536}{5}\)
Vậy \(A=\frac{1536}{5}\)
T
22 tháng 2 2020
C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)
C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)
C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)
C = \(\frac{-233}{135}\)
D = \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)
D = \(-4.\frac{12}{13}\)
D = \(\frac{-48}{13}\)
E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)
E = \(5.4-4.3+5-0,3.20\)
E = \(20-12+5-6\)
E = \(8+\left(-1\right)\)
E = \(7\)
F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\)
F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)
F = \(\frac{-11}{12}\)
Chúc cậu hk tốt ~
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)
b)
\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)
10 tháng 2 2020
\(a,\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)
\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)
\(=\frac{31}{25}:\left(-\frac{29}{25}\right)\)
\(=\frac{31}{25}.\frac{-25}{29}\)
\(=-\frac{31}{29}\)
\(b,\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)
\(=\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}.\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}.\frac{19}{4}\right)\)
\(=\left(\frac{109}{6}-\frac{13}{10}\right):\frac{19}{3}\)
\(=\frac{253}{15}.\frac{3}{19}\)
\(=\frac{253}{95}\)
Số to :v
18 tháng 12 2016
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
\(C=\left(0,5\right)^{-4}-625^{0,25}-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}+19\left(-3\right)^{-3}=\left(2^{-1}\right)^{-4}-\left(5^4\right)^{\frac{1}{4}}-\left[\left(\frac{3}{2}\right)^2\right]^{-\frac{3}{2}}+19.\frac{1}{\left(-3\right)^3}\)
\(=2^4-5-\left(\frac{3}{2}\right)^{-3}-\frac{19}{27}\)
\(=11-\left(\frac{2}{3}\right)^3-\frac{19}{27}=10\)
\(C=\left(0,5\right)^{-4}-625^{0,25}-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}+19.\left(-3\right)^{-3}\)
\(=\left(\frac{1}{2}\right)^{-4}-625^{\frac{1}{4}}-\left(\frac{9}{4}\right)^{-\frac{3}{2}}+19.\left(-3\right)^{-3}\)
\(=2^4-\sqrt[4]{625}-\left(\frac{4}{9}\right)^{\frac{3}{2}}+19.\left(\frac{1}{\left(-3\right)^3}\right)\)
=\(16-5-\sqrt[2]{\left(\frac{4}{9}\right)^3}+19.\frac{1}{-27}=11-\frac{8}{27}-\frac{19}{27}=10\)