tính:
\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
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
29 tháng 9 2016
A = (1 - 2/3 + 4/3) - (4/5 - 1) + (7/5 + 2)
A= (3/3 - 2/3 + 4/3) - (4/5 - 5/5) + (7/5 + 10/5)
A= 5/3 + 1/5 + 17/5
A= 5/3 +18/5
A= 25/15 + 54/15
A= 79/15
B= (-3 + 3/4 - 1/3 ) : (5 + 2/5 - 2/3)
B= (-36/12 + 9/12 - 4/12) : (75/15 + 6/15 - 10/15)
B= -31/12 : 71/15
B= -155/284
C= (3/5 - 4/5 ) . (2/7 - 3/14) - (5/9 - 7/27) . (1 - 3/5) + (1 - 11/12) . (1-11/12)
C= -1/5 . 1/14 - 8/27 . 2/5 + 1/12 . 1/12
C=-1/70 - 16/135 + 1/144
C=-216/15120 - 1792/15120 + 105/15120
C=-1903/15120
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\frac{3}{7}.\left( { - \frac{1}{9}} \right) + \frac{3}{7}.\left( { - \frac{2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} + \frac{-2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} - \frac{6}{9}} \right)\\ = \frac{3}{7}.\frac{{ - 7}}{9} = \frac{{ - 1}}{3}\end{array}\)
b)
\(\begin{array}{l}\left( {\frac{{ - 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ - 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.\left( {\frac{5}{{12}} + \frac{7}{{12}}} \right) + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.1 + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 13}}{{13}}\\ = -1\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} - \frac{1}{3}} \right):\frac{5}{9}\\ = \left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right].\frac{9}{5} + \left( {\frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left( {\frac{{ - 2}}{3} + \frac{3}{7} + \frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left[ {\left( {\frac{{ - 2}}{3} - \frac{1}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right].\frac{9}{5}\\ = \left( { - 1 + 1} \right).\frac{9}{5}\\ = 0.\frac{9}{5} = 0\end{array}\)
d)
\(\begin{array}{l}\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{2}{3}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{{10}}{{15}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 9}}{15}\\= \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 3}}{5}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{5}{9}.\frac{{ - 5}}{3}\\ = \frac{5}{9}.\left( {\frac{{ - 22}}{3} - \frac{5}{3}} \right)\\ = \frac{5}{9}.\frac{-27}{3}= \frac{5}{9}.\left( { - 9} \right) = - 5\end{array}\)
e)
\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} - \left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{{ - 2}}{{97}}} \right) - \frac{1}{{35}} - \frac{3}{4} + \left( {\frac{{ - 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} - \frac{2}{{97}} - \frac{1}{{35}} - \frac{3}{4} - \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} - \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} - \frac{3}{4} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} - \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} - \frac{{33}}{{44}} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { - 1} \right) - \frac{2}{{97}}\\ = - \frac{2}{{97}}\end{array}\)
EC
8 tháng 9 2016
Câu 1:
a)\(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(-\frac{9}{2}\right)\right]-\frac{5}{6}\)
\(=\frac{3}{4}-\frac{1}{4}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)
\(=\frac{1}{2}-\frac{4}{3}\)
\(=-\frac{5}{6}\)
b)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
\(=5\)
EC
8 tháng 9 2016
Câu 2:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
Vậy -1\(\le\)x<7
2 tháng 2 2022
\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)
\(=\dfrac{880}{20\cdot4}=11\)
\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)
\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)
= 7+(37/12)-(1/12+5)
=7+(37/12)-6
=49/12
\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{7}{12}-\frac{1}{2}+3-\frac{1}{12}-5\)
\(=\left(7+3-5\right)+\left(\frac{7}{12}-\frac{1}{12}-\frac{1}{2}\right)\)
\(=5+\left(\frac{6}{12}-\frac{1}{2}\right)\)
\(=5+0\)
\(=5\)