tính giá trị biểu thức (tính nhanh nếu có thể)
a,\(\left(\frac{7}{4}.\frac{-4}{5}+\frac{7}{2}.\frac{-1}{5}\right).50\%-0,1\)
b,\(\left(2\frac{1}{3}+3\frac{1}{2}\right).0,2+25\%\)
Muốn tick thì nhanh lên
tính giá trị biểu thức(tính nhanh nếu có thể)
a,\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
b,\(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
Nhanh thì tick
tính giá trị biểu thức (tính nhanh nếu có thể)
a,\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
b,\(\frac{2}{3}+\frac{1}{3}.\left(-\frac{4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
Tính giá trị của biểu thức sau (tính hợp lí, nếu có thể):
a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\);
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} = \frac{{ -29}}{{35}}\end{array}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)
\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ = (- 2).45,3 = - 90,6\end{array}\)
a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)
b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)
\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)
c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)
d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)
tính giá trị biểu thức (tính nhanh nếu có thể)
a,\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
b,\(\frac{2}{3}+\frac{1}{3}.\left(-\frac{4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
Tính giá trị biểu thức(giút gọn biểu thức)
A=\(\left(\left(\frac{2}{193}-\frac{3}{386}\right)\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{2001}+\frac{11}{4002}\right)\cdot\frac{2001}{25}+\frac{9}{2}\right)\)
\(B=\left(1+2+3+4+.....+100\right)\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)\cdot\left(\frac{6}{3}\cdot12-2,1\cdot3,6\right)\)
C=\(\frac{2\cdot8^4\cdot27^2+4\cdot69}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(F=1-\frac{1}{1+\frac{2}{1-\frac{3}{1-4}}}\)
ai làm đúng nhanh dễ hiểu thì mk tick cho
Cho biểu thức: \(A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\)
Hãy tính giá trị của A theo hai cách:
a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.
b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
1/ Tính giá trị biểu thức
A=\(\frac{7}{35}+\left(-1\frac{3}{4}+\frac{12}{7}\right)-\left(\frac{1}{4}-\frac{2}{7}-\frac{13}{35}\right)-\frac{3}{7}\)\(\frac{3}{7}\)
B=\(\sqrt{\frac{25}{16}-\frac{\left(-3\right)^2}{\left|-4\right|}}+\frac{\sqrt{\left(-7\right)^2}}{4}-3.\left(\frac{5}{2}\right)^2\)
Tính giá trị biểu thức
a)\(\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right)\):(-2)
b)\(\left(\frac{2}{25}-1,008\right):\frac{4}{7}:[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\)
Tính giá trị của mỗi biểu thức sau:
a) \(0,2 + 2,5:\frac{7}{2}\)
b) \(9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\)
a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)
b)
\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)