Tính nhanh biểu thức sau:
\(A=\left(7-\frac{3}{4}+\frac{1}{3}\right)-\left(6+\frac{5}{4}-\frac{4}{3}\right)-\left(5-\frac{7}{4}+\frac{5}{3}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
NN
16 tháng 7 2016
\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)
\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)
\(P=\frac{37}{60}\)
\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)
\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)
\(Q=\frac{1044}{4375}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(\begin{array}{l}A = \left( {7 - \frac{2}{5} + \frac{1}{3}} \right) - \left( {6 - \frac{4}{3} + \frac{6}{5}} \right) - \left( {2 - \frac{8}{5} + \frac{5}{3}} \right)\\A = 7 - \frac{2}{5} + \frac{1}{3} - 6 + \frac{4}{3} - \frac{6}{5} - 2 + \frac{8}{5} - \frac{5}{3}\\A = \left( {7 - 6 - 2} \right) + \left( { - \frac{2}{5} - \frac{6}{5} + \frac{8}{5}} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
Chú ý:
Trong phép tính chỉ có phép cộng trừ, ta có thể đổi chỗ các số hạng tùy ý kèm theo dấu của chúng.
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
\(A=7-\frac{3}{4}+\frac{1}{3}-6-\frac{5}{4}+\frac{4}{3}-5+\frac{7}{4}-\frac{5}{3}\)
\(A=7-6-5+\left(\frac{-3}{4}-\frac{5}{4}+\frac{7}{4}\right)+\left(\frac{1}{3}+\frac{4}{3}-\frac{5}{3}\right)\)
\(A=-4-\frac{1}{4}\)
\(A=\frac{-17}{4}\)
#)Giải :
\(A=\left(7-\frac{3}{4}+\frac{1}{3}\right)-\left(6+\frac{5}{4}-\frac{4}{3}\right)-\left(5-\frac{7}{4}+\frac{5}{3}\right)\)
\(A=7-\frac{3}{4}+\frac{1}{3}-6-\frac{5}{4}+\frac{4}{3}-5-\frac{7}{4}+\frac{5}{3}\)
\(A=\left(7-6-5\right)-\left(-\frac{3}{4}-\frac{5}{4}+\frac{7}{4}\right)+\left(\frac{1}{3}+\frac{4}{3}-\frac{5}{3}\right)\)
\(A=-4-\frac{1}{4}+0\)
\(A=-4\frac{1}{4}=-\frac{17}{4}\)
#~Will~be~Pens~#