Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
yutyugubhujyikiu
bài 1 : tính
a)\(\frac{-5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\) b) \(\left(\frac{3}{9}-\frac{9}{18}\right)+\frac{3}{6}-\left(\frac{1}{3}-\frac{1}{2}\right)-\frac{5}{15}\) c) \(\frac{9}{18}+\frac{16}{32}-\frac{12}{46}-\frac{9}{17}\) d) \(\left(\frac{14}{18}+\frac{-16}{27}\right)-\left(\frac{2}{3}-\frac{5}{15}\right)\)
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
Kiểm tra bài : Nhân, chia số hữu tỉ
Thực hiện phép tính :
(1) \(-\frac{3}{2}.\frac{7}{10}=\frac{-3.7}{2.10}=\frac{-21}{20}\)
(2) \(\frac{-5}{3}.\frac{6}{11}=\frac{-5.6}{3.11}=\frac{-30}{33}\)
(3) \(2\frac{1}{3}.\left(-1\frac{2}{3}\right)=\frac{7}{3}.\left(-\frac{5}{3}\right)=\frac{7.\left(-5\right)}{3.3}=-\frac{35}{9}\)
(4) \(\frac{9}{10}:\left(-\frac{15}{11}\right)=\frac{9}{10}.\left(\frac{-11}{15}\right)=\frac{9.\left(-11\right)}{10.15}=-\frac{99}{150}=-\frac{33}{50}\)
(5) \(\left(-1\right):\frac{3}{8}=\frac{\left(-1\right).8}{3}=-\frac{8}{3}\)
(6) \(\frac{1}{2}.\left(-\frac{5}{4}\right).\frac{8}{7}=\frac{1.\left(-5\right)}{2.4}.\frac{8}{7}=-\frac{5}{8}.\frac{8}{7}=-\frac{5.8}{8.7}=-\frac{5}{7}\)
(7) \(\frac{-9}{2}.\frac{2}{18}.\frac{1}{7}=\left(-\frac{9}{2}.\frac{2}{18}\right).\frac{1}{7}=\left(-\frac{9.2}{2.18}\right).\frac{1}{7}=-\frac{18}{36}.\frac{1}{7}=-\frac{18.1}{36.7}=-\frac{1}{14}\)
(8) \(\left(\frac{9}{2}-\frac{1}{3}\right).\frac{6}{17}=\left(\frac{27}{6}-\frac{2}{6}\right).\frac{6}{17}=\frac{27-2}{6}.\frac{6}{17}=\frac{25}{6}.\frac{6}{17}=\frac{25.6}{6.17}=\frac{25}{17}\)
(9) \(\left(-\frac{12}{13}:\frac{36}{39}\right).\frac{3}{5}=\left(-\frac{12}{13}.\frac{39}{36}\right).\frac{3}{5}=\left(\frac{-12.39}{13.36}\right).\frac{3}{5}=-\frac{1.3}{5}=-\frac{3}{5}\)
(10) \(\left(-\frac{3}{7}+\frac{7}{9}\right):\frac{4}{7}+\left(-\frac{4}{7}+\frac{2}{9}\right):\frac{4}{7}=\left(\left(-\frac{3}{7}+\frac{7}{9}\right)+\left(-\frac{4}{7}+\frac{2}{9}\right)\right):\frac{4}{7}\)
\(=\left(\left(-\frac{27}{63}+\frac{49}{63}\right)+\left(-\frac{36}{63}+\frac{14}{63}\right)\right):\frac{4}{7}=\left(\left(-\frac{27+49}{63}\right)+\left(\frac{-36+14}{63}\right)\right):\frac{4}{7}\)
\(=\left(\left(\frac{22}{63}\right)+\left(-\frac{22}{63}\right)\right):\frac{4}{7}\)
\(=\frac{22+\left(-22\right)}{63}:\frac{4}{7}=\frac{0}{63}:\frac{4}{7}=0\)
Mình đăng các bài toán này lên thứ nhất là để kiểm tra năng lực thứ hai các bạn có thể xem đây và rút ra lời giải cho các bài khác và nếu mình sai chỗ nào các bạn chỉ mình sẽ chỉnh
Thực hiện phép tính:
a,\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
b,\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
c,\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)
\(=10:\left(\frac{-2}{3}\right)\)
\(=-15\)
b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\frac{-8}{27}+\frac{1}{10}\)
\(=\frac{-8}{3}+\frac{1}{10}\)
\(=\frac{-77}{30}\)
c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)
\(=\frac{-1}{3}\)
\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)
\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)
\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)
\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)
\(=-15\)
\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)
\(=-\frac{8}{3}+\frac{1}{10}\)
\(=-\frac{77}{30}\)
\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)
\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)
\(=\frac{1}{3}\)
\(A=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)
\(B=3-\frac{2}{3}+\frac{3}{5}\left(-\frac{10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(\left(3-\frac{3}{4}+\frac{2}{3}\right)-\left(2+\frac{4}{3}-\frac{3}{2}\right)-\left(1-\frac{7}{3}-\frac{9}{2}\right)\)
\(D=\left[-\frac{54}{64}-\left(\frac{1}{9}:\frac{8}{27}\right):-\frac{1}{3}\right]:-\frac{81}{128}\)
\(-\frac{3}{5}+\frac{2}{61}-\frac{17}{51}\)
\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)+\left(\frac{3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
\(\frac{1}{4}-3\cdot\left(\frac{1}{12}+\frac{3}{8}\right)\)
\(\left(-\frac{2}{3}+\frac{3}{5}\right):\frac{1}{50}-30\)
\(\left(-\frac{4}{5}+\frac{5}{7}\right):\frac{3}{17}+\left(-\frac{1}{5}+\frac{2}{7}\right):\frac{3}{17}\)
\(\frac{1}{3}-\frac{1}{7}+\frac{8}{9}+\frac{15}{12}-\frac{3}{5}-\frac{2}{3}\)
► Các cậu giúp mình nha. Mình sẽ tick cho ai có đáp án nhanh nhất và đúng nhất nhé. Mình cảm ơn tất cả các cậu trước ạ. Kamsa~
Những câu từ D trở đi là các câu riêng biệt ak bạn
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(B=3-{2\over3}+{3\over5}(-{10\over9}-{25\over3})-{5\over6}\)\(={7\over3}+{3\over5}(-{10\over9}-{75\over9})-{5\over6}\)\(={14\over6}-{5\over6}+{3\over5}({-85\over9})\)\(={3\over2}-{17\over3}={9\over6}-{34\over6}={-25\over6}\)
\(\frac{3}{5}\left(\frac{5}{6}-\frac{10}{9}\right):\frac{1}{2}-\frac{9}{5}:\frac{18}{15}+\frac{28}{15}:\left(\frac{-7}{10}\right)\)
\(3\left(\frac{-2}{3}\right)^2-2\left(\frac{-1}{2}\right)^3-\left(-\frac{3}{4}\right)^0\)
Tính từng bước các phép tính trên
Tính:
A=\(\left(1-\frac{2}{3}+\frac{4}{3}\right)-\left(\frac{4}{5}-1\right)+\left(\frac{7}{5}+2\right)\)
B=\(\left(-3+\frac{3}{4}-\frac{1}{3}\right):\left(5+\frac{2}{5}-\frac{2}{3}\right)\)
C=\(\left(\frac{3}{5}-\frac{4}{15}\right).\left(\frac{2}{7}-\frac{3}{14}\right)-\left(\frac{5}{9}-\frac{7}{27}\right)\)\(.\left(1-\frac{3}{5}\right)+\left(1-\frac{11}{12}\right).\left(1+\frac{11}{12}\right)\)
D=\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{-5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{4}{3}}\)
1/ \(-\frac{4}{5}:\left(\frac{20}{9}-\frac{8}{15}\right)+\frac{3}{5}\left|\frac{10}{-9}\right|-\frac{1}{2}\left(\frac{4}{3}-\frac{2}{5}\right)\)
a)\(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right) \)
b)\(\frac{1}{2}\sqrt[]{64-\sqrt{\frac{4}{25}+1^{2012}}}\)
c) \(9.\left(\frac{1}{3}\right)^3:\left[\left(\frac{-2}{3}\right)^2+0,5-1\frac{1}{2}\right]\)
d)\(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
