Tính hợp lí: \(\frac{4}{5}.50\frac{1}{2}+\frac{2}{3}.41\frac{7}{5}-\frac{5}{4}.40\frac{1}{2}+\frac{2}{3}.\frac{2}{7}\)
tính hợp lí
1, \(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7-\frac{2}{3}+\frac{6}{7}}\)
2, \(5,25+3\frac{4}{9}+2\frac{1}{27}+4,75+3\frac{1}{4}+2+\frac{3}{4}\)
3, \(12\frac{3}{5}+\frac{17}{24}+\frac{7}{5}\)
4, \(4\frac{2}{3}+2\frac{1}{5}+2\frac{2}{3}+3\frac{3}{5}\)
Tính hợp lí:
A =\((\frac{2}{7}\times\frac{1}{4}-\frac{1}{3}\times\frac{2}{7})\div(\frac{2}{7}\times\frac{3}{9}-\frac{2}{7}\times\frac{2}{5})\)
B = \(\frac{(\frac{1}{5}-\frac{2}{7})\times\frac{3}{4}-\frac{3}{4}\times(\frac{1}{3}-\frac{2}{7})}{\frac{1}{5}\times\frac{2}{7}-\frac{1}{3}\times(\frac{2}{7}+\frac{3}{9})+\frac{3}{9}\times\frac{1}{5}}\)
CÓ LỜI GIẢI THÍCH CHI TIETS NHÉ AI NHANH MK TICK
A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
=(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)
Tính hợp lí: \(A=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
Đáp số là 1 vì cả hai phân số đều bằng 2/7 và 2/7:2/7=14/14=1.Đúng 100000000000000000000% luôn bạn ơi!
đưa tử về 1 thử coi bn
trước đây cô giáo cũng bày mk thế
Tính 1 cách hợp lí nhất :
\(\frac{-2}{5}.\frac{4}{7}+\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}\)
\(\frac{-3}{5}.\frac{4}{7}+\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}\)
\(=\frac{-3}{5}.\frac{4}{7}+\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}.\frac{1}{7}\)
\(=\frac{-3}{5}\left(\frac{4}{7}+\frac{2}{7}+\frac{1}{7}\right)\)
\(=\frac{-3}{5}.1\)
\(=\frac{-3}{5}\)
\(-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}x\frac{2}{7}+-\frac{3}{5}=-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}\left(\frac{2}{7}+1\right)=-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}x\frac{9}{7}.\)
\(-\frac{2}{35}-\frac{27}{35}=-\frac{29}{35}\)
tôi tính nhẩm vậy ko biết có đứng ko
\(A=\frac{3}{7}-\frac{3}{17}+\frac{3}{37}:\frac{5}{7}-\frac{5}{17}+\frac{5}{37}\) + \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}:\frac{7}{5}-\frac{7}{4}-\frac{7}{3}-\frac{7}{2}\)
Tính hợp lí nha:
(Dấu : là phần nha)
Làm ơn giúp mình nha
Thực hiện phép tính hợp lí nếu có thể:
a/ \(\frac{6}{7}+\frac{1}{7}\cdot\frac{2}{7}+\frac{1}{7}\cdot\frac{5}{7}\)
b/\(\frac{2}{3}\cdot\frac{5}{7}\cdot\frac{-3}{8}\cdot\frac{11}{5}\)
c/\(11\frac{4}{7}-\left(2\frac{3}{5}+5\frac{4}{7}\right)\)
d/\(\frac{3}{4}-\frac{3}{4}\cdot\left(\frac{2}{3}+1\right)\)
e/\(0.5\cdot1\frac{1}{3}\cdot75\%:\frac{2}{5}+\frac{3}{5}\)
f/\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\cdot\left(-2\right)^2\)
g/\(1\frac{3}{8}+\left(\frac{-5}{6}+\frac{7}{12}\right):\frac{2}{3}\)
h/\(1\frac{1}{4}\cdot\frac{-3}{2}+50\%\cdot98\)
i/\(\left(2,09:1,1+4,5\right)\cdot\frac{5}{8}+4,32\)
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
trời ơi bài dễ thế này tự làm đi còn hỏi
Đã biết thì đã không hỏi. Đồ kazuto kirigaya xấu tính!
tình hợp lí:
\(\frac{\left(\frac{1}{5}-\frac{2}{7}\right)\cdot\frac{3}{4}-\frac{3}{4}\cdot\left(\frac{1}{3}-\frac{2}{7}\right)}{\frac{1}{5}\cdot\frac{2}{7}-\frac{1}{3}\cdot\left(\frac{2}{7}+\frac{3}{9}\right)+\frac{3}{9}+\frac{1}{5}}\)
Tính hợp lý \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)
\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}+\frac{1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}\)
\(=\frac{2}{3}+\frac{1}{3}\)
\(=1\)
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: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)