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Thực hiện phép tính ( bằng cách hợp lí nếu có thể )
A. 2/3 - (- 1/4)+3/5-(+7/45)-(-5/9)+1/12+1/39
B. 6.(-2/3)^2 - 3(-2/3)^2 - 2 : (-3/2)+4
\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
\(=\left(-8\right)+2+1+\frac{-3}{10}\)
\(=\left(-5\right)+\frac{-3}{10}=\frac{-53}{10}=-5,3\)
(3-1/4+2/3)-(5-1/3-6/5)-(6-7/4+3/2)
=3-1/4+2/3-5-1/3-6/5-6-7/4+3/2
=(3-5-6)+(-1/4-7/4)+(2/3-1/3)+(6/5/3/2
=(-8)+2+1+(-3/10)
=(-5)+(-3/10)=-53/10=-5,3
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
\(\frac{2}{3}-\left(-\frac{1}{4}\right)+\frac{3}{5}-\frac{7}{45}-\left(-\frac{7}{9}\right)+\frac{1}{12}+\frac{1}{39}\)
\(=\frac{2}{3}+\frac{1}{4}+\frac{3}{5}-\frac{7}{45}+\frac{7}{9}+\frac{1}{12}+\frac{1}{39}\)
\(=\left(\frac{2}{3}+\frac{1}{4}+\frac{1}{12}\right)+\left(\frac{7}{9}+\frac{3}{5}-\frac{7}{45}\right)+\frac{1}{39}\)
\(=\left(\frac{8}{12}+\frac{3}{12}+\frac{1}{12}\right)+\left(\frac{35}{45}+\frac{27}{45}-\frac{7}{45}\right)+\frac{1}{39}\)
\(=\frac{12}{12}+\frac{55}{45}+\frac{1}{39}=1+\frac{11}{9}+\frac{1}{39}=\frac{351}{351}+\frac{429}{351}+\frac{9}{351}=\frac{789}{351}\)
2 3 − ( − 1 4 ) + 3 5 − 7 45 − ( − 7 9 ) + 1 12 + 1 39 = 2 3 + 1 4 + 3 5 − 7 45 + 7 9 + 1 12 + 1 39 = ( 2 3 + 1 4 + 1 12 ) + ( 7 9 + 3 5 − 7 45 ) + 1 39 = ( 8 12 + 3 12 + 1 12 ) + ( 35 45 + 27 45 − 7 45 ) + 1 39 = 12 12 + 55 45 + 1 39 = 1 + 11 9 + 1 39 = 351 351 + 429 351 + 9 351 = 789 351
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)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
\(=5\)
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
a) \(-1\frac{5}{7}.15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{3}-\frac{4}{5}+\frac{1}{7}\right)\)
\(=1\frac{5}{7}.\left(-15\right)+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{70}{105}-\frac{84}{105}+\frac{15}{105}\right)\)
\(=\left(-15\right)\left(1+\frac{5}{7}+\frac{2}{7}\right)+\left(-105\right).\frac{1}{105}\)
\(=-30-1=-31\)
b) \(\frac{2}{3}+\frac{3}{4}.\left(-\frac{4}{9}\right)\)
= \(=\frac{2}{3}+\frac{3.\left(-4\right)}{4.9}=\frac{2}{3}+\frac{-1}{3}=\frac{1}{3}\)
c) \(\left(\frac{3}{4}-0,2\right).\left(0,4-\frac{4}{5}\right)\)
\(=\left(\frac{3}{4}-\frac{1}{5}\right).\left(\frac{2}{5}-\frac{4}{5}\right)=\frac{11}{20}.\left(\frac{-2}{5}\right)=\frac{11.\left(-1\right).2}{2.10.5}=\frac{-11}{50}\)
\(12.\left(-\frac{2}{3}\right)+\frac{4}{3}\)
\(=-8+\frac{4}{3}\)
\(=\frac{-20}{3}\)
\(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
\(=\frac{4}{5}.\frac{225}{16}\)
\(=\frac{45}{4}\)
\(12.\left(-\frac{2}{3}\right)+\frac{4}{3}\)
\(=-8+\frac{4}{3}=-\frac{20}{3}\)( bấm máy tính ... số cx ko to lắm nên cx hợp lí :v)
\(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
\(=\frac{4}{5}.\frac{225}{16}\)
rút gọn ta đc \(=\frac{45}{4}\)