câu 1 tính hợp lý nếu có thể
\(a,\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
giúp mik với !!!
Tính hợp lý nếu có thể:
a) \(2\frac{1}{7}:\left(\frac{1}{15}-1\frac{2}{5}\right)-2\frac{1}{7}:\left(\frac{17}{15}+2\frac{1}{5}\right)\)
b) \(\left(\frac{-2}{3}\right)^{-4}.\left(-4\right)^2-\left[\left(\frac{7}{3}\right)^0\right]^{2016}-10\frac{1}{3}\)
a) A = \(\frac{15}{7}:\left(\frac{1}{15}-\frac{7}{5}\right)-\frac{15}{7}:\left(\frac{17}{15}+\frac{11}{5}\right)=\frac{15}{7}:\frac{-20}{15}-\frac{15}{7}:\frac{50}{15}\)
A = \(\frac{15}{7}.\frac{15}{-20}-\frac{15}{7}.\frac{15}{50}=\frac{15}{7}.\left(\frac{-15}{20}-\frac{15}{50}\right)=\frac{15}{7}.\frac{-105}{100}=-\frac{9}{4}\)
b) B = \(\frac{1}{\left(-\frac{2}{3}\right)^4}.\left(-4\right)^2-1^{2016}-10\frac{1}{3}=\frac{1}{\frac{16}{81}}.16-1-10\frac{1}{3}=\frac{81}{16}.16-1-10\frac{1}{3}\)
B = \(81-1-10-\frac{1}{3}=70-\frac{1}{3}=\frac{209}{3}\)
câu 1
\(a,-8\frac{2}{5}:\left(-2\frac{4}{5}\right)\)
câu 2 tính hợp lý nếu có thể
\(a,15-2\frac{1}{3}:\left(\frac{4}{9}-\frac{1}{6}\right)\)
giúp mik với nhé !!!
Câu 1:
a) \(=\frac{42}{5}:\frac{14}{5}\)
\(=3\)
b)
\(=15-\frac{7}{3}:\left(\frac{4}{9}-\frac{1}{6}\right)\)
\(=15-\frac{7}{3}:\frac{4}{9}+\frac{7}{3}:\frac{1}{6}\)
\(=15-\frac{21}{4}+14\)
\(=29-\frac{21}{4}=29-5-\frac{1}{4}\)
\(=24-\frac{1}{4}=\frac{95}{4}\)
Thực hiện phép tính (hợp lí nếu có thể):
\(1,\frac{-1}{3}-\frac{-3}{5}-\frac{1}{6}+\frac{1}{43}-\frac{-3}{7}+\frac{-1}{2}-\frac{1}{35}\\ \\ 2,\left(-\frac{1}{3}+\frac{7}{13}\right)-\left(\frac{-16}{24}+\frac{6}{26}+\frac{9}{13}\right)\)\(3,\frac{-7}{3}-\left[\frac{2}{5}-\left(\frac{1}{3}+\frac{-5}{25}\right)\right]\\ 4,\left(2\frac{1}{4}-3\frac{1}{5}\right)-\left[\frac{-3}{4}+\left(\frac{4}{5}-2019\right)\right]\)
BÀi 1: Thực hiện phép tính ( tính nhanh nếu có thể)
a.\(\left(-\frac{1}{2}\right)-\left(-\frac{3}{5}\right)+\left(-\frac{1}{9}\right)+\frac{1}{71}-\left(-\frac{2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
b.\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
c.\(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{1}{3}-1\frac{1}{15}\right)\)
Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)
a)\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b)\(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
C)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{17}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
d)\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
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 hợp lý nếu có thể:
a, \(\frac{9\times\left(-3\right)^6-5\times3^7}{2^4\times\left(-3\right)^7}\)
b,\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
Giải chi tiết giúp mk bài này với: Tính hợp lý :
\(A=\left(-\frac{2}{9}\right)-\left(-\frac{1}{121}\right)+\left(-\frac{3}{5}\right)+\left(-\frac{1}{2}\right)-\left(-1\frac{13}{18}\right)-\frac{1}{7}-\frac{9}{35}\)
\(B=0,75+\frac{3}{5}+\left(\frac{2}{9}-1\frac{3}{5}+1\frac{1}{4}\right)\)
\(C=1-2+3-4+...+79-80\)
c) C = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 79 - 80 )
C = ( -1 ) + ( -1 ) + ... + ( -1 )
C = ( -1 ) x ( 80 - 1 + 1 ) : 2
C = ( -1 ) x 80 : 2
C = ( -40 )
C=1-2+3-4+...+79-80
=(1-2)+(3-4)+...+(79-80)
=-1+(-1)+...+(-1) (có 80 số hạng bàng -1)
=-1*80
=-80
Tính bằng cách hợp lí ( Nếu có thể )
a) \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}\)
b) \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)\)
a)\(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{117}{12}-\frac{2}{12}=\frac{115}{12}\)
b)\(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)
a. \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)
b. \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-1\frac{1}{2}\)
= \(\frac{13}{4}.1-\frac{3}{2}=\frac{13}{4}-\frac{3}{2}=\frac{7}{4}\)
c. \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2004}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}=\frac{1}{2004}\)
Giải :
a) \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)
b) \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)
c) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)