(0,4)2+\(5\frac{1}{2}\): (1,9-4,1)+\(\frac{2^3}{\left(-2\right)^2}\)
\(A=\left[\left(2,1^2-4,01\right)^2.2,5+7,3\right]\div\left[1,9-0,4^2-\left(\frac{4}{5}\right)^2\right]\)
I,Rút gọn biểu thức chứa dấu GTTĐ
1.A= |x-3,5|+|4,1-x|
2.B= |-x+3,5|+|x-4,1|
II,Rút gon BT khi \(\frac{-3}{5}< x< \frac{1}{7}\)
\(1,A=\left|x-\frac{1}{7}\right|-\left|x+\frac{3}{5}\right|+\frac{4}{5}\)
\(2,B=\left|-x+\frac{1}{7}\right|+\left|-x-\frac{3}{5}\right|-\frac{2}{6}\)
Tính:
a)\(0,75 - \frac{5}{6} + 1\frac{1}{2};\)
b)\(\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right);\)
c)\(0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\)
d)\(\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right);\)
e) \(\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5};\)
g)\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right)\)
a)
\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)
c)
\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)
d)
\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)
e)
\(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)
g)
\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)
Thay số thích hợp thay vào dấu “?” trong các câu sau:
a)\({\left[ {{{\left( {\frac{{ - 2}}{3}} \right)}^2}} \right]^5} = {\left( {\frac{{ - 2}}{3}} \right)^?};\) b)\({\left[ {{{\left( {0,4} \right)}^3}} \right]^3} = {\left( {0,4} \right)^?}\) c)\({\left[ {{{\left( {7,31} \right)}^3}} \right]^0} = ?\)
a)\({\left[ {{{\left( {\frac{{ - 2}}{3}} \right)}^2}} \right]^5} = {\left( {\frac{{ - 2}}{3}} \right)^{2.5}} = {\left( {\frac{{ - 2}}{3}} \right)^{10}}\)
Vậy dấu “?” bằng 10.
b) \({\left[ {{{\left( {0,4} \right)}^3}} \right]^3} = {\left( {0,4} \right)^{3.3}} = {\left( {0,4} \right)^9}\)
Vậy dấu “?” bằng 9.
c) \({\left[ {{{\left( {7,31} \right)}^3}} \right]^0} = 1\)
Vậy dấu “?” bằng 1.
Tính:
a)\(\left\{\left[\left(6,2:0,31-\frac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\frac{4}{11}:0,1\right).\frac{1}{33}\right]\)
b)\(0,4\left(3\right)+0,6\left(2\right)-2\frac{1}{2}.\left[\left(\frac{1}{2}+\frac{1}{3}:0,5\left(8\right)\right)\right]:\frac{50}{53}\)
c)\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
a)\(\left(\frac{1}{13}\right)^{13}:\left(\frac{1}{3}\right)^{x-2}=\frac{1}{81}\)
b)\(\left(0,4\right)^{x-1}:\left(\frac{2}{5}\right)^2=\frac{8}{125}\)
b) ( 2/5 )^ x-1 : ( 2/5 )^2 = ( 2/5 )^3
(2/5)^ x-1 = ( 2/5) ^3 . (2/5)^2
(2/5)^ x-1 = ( 2/5 ) ^ 6
=> x-1 = 6
=> x= 7
Rút gọn
1.\(\left(\frac{2}{45}-\frac{4}{13}-\frac{1}{3}\right):\left(\frac{3}{13}-\frac{4}{15}+\frac{2}{13}\right)\)
2.\(\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{15}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)2\frac{2}{17}}\)
3.\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)
3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
=2/7-2/7=0
Tính:
a) \(3^{-2}.\left(\frac{2}{3}\right)^{-4}.\left(-1\frac{1}{2}\right)^{-3}\)
b) \(0,02^{-3}.10^{-4}.\left(\frac{4}{5}\right)^{-2}.\left(0,4\right)^4\)
c) \(\frac{2^{-2}-\left(\frac{3}{4}\right)^{-4}.\left(-\frac{1}{2}\right)^2}{10^{-1}+\left(-\frac{1}{8}\right)^0}\)
Tính :
a) \(1\frac{7}{20}:2,7+2,7:1,35+\left(0,4:2\frac{1}{2}\right).\left(4,8-1\frac{3}{40}\right)\)
b) \(\left[\left(6\frac{3}{5}-3\frac{3}{14}\right):5\frac{5}{6}\right]:\left[\left(21-1,25\right):2,5\right]\)