Tỉ số giữa A và B biết
\(A=\frac{7!.4!}{10!}.\left(\frac{8!}{3!.5!}-\frac{9!}{2!.5!}\right)\)
\(B=\frac{2}{\left(-3\right)^4}\)
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,\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{71}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{8}\)
d,\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
e,\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(\frac{-2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
g,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)
a, \(\left(7+\frac{7}{5}-\frac{2}{3}\right)-\left(4+\frac{4}{5}+\frac{3}{8}\right)+\left(3-\frac{3}{5}+\frac{2}{3}+\frac{3}{8}\right)\)
b.\(-\frac{13}{25}.\frac{5}{32}.\left(\frac{25}{-13}\right).\left(-64\right)\)
c.\(\frac{7}{80}:\left(-1\frac{3}{4}\right)-\frac{2}{9}:\left(8-\frac{8}{3}\right)-\left(-\frac{5}{24}\right)\left(-\frac{10}{3}+2\frac{8}{15}\right)\)
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
1.Thực hiện phép tính:(hepl me)
a)\(4\frac{5}{9}:\left(\frac{-5}{7}\right)+\frac{49}{9}:\left(\frac{-5}{7}\right)\)
b)\(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
c)\(\left(\frac{3}{4}\right)^4\times\left(\frac{8}{9}\right)^2\)
d)\(\left(\frac{-3}{5}\right)^6\times\left(-\frac{5}{3}\right)^5\)
e)\(\frac{8^{14}}{4^4\times64^5}\)
f)\(\frac{9^{10}\times27^7}{81^7\times3^{15}}\)
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
Tính:
a)\(\frac{3}{5}+\left(\frac{-4}{3}\right)+\left(\frac{-3}{4}\right)\)
b) \(\frac{5}{8}-\left(\frac{-2}{5}\right)-\frac{3}{10}\)
c) \(\left(\frac{-7}{2}\right)+\left(\frac{-2}{3}\right)+\left(\frac{-3}{5}\right)\)
d) \(\frac{3}{4}-[\left(\frac{-5}{3}\right)-\left(\frac{1}{12}+\frac{2}{9}\right)]\)
\(A=\left(\frac{5}{3}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)
\(A=\left(\frac{5}{3}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)
\(A=\frac{5}{3}-\frac{3}{7}+9-2-\frac{5}{7}+\frac{2}{3}+\frac{8}{7}-\frac{4}{3}-10\)
\(A=\left(\frac{5}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{7}+\frac{5}{7}-\frac{8}{7}\right)+\left(9-2-10\right)\)
\(A=1-1-3\)
\(A=-3\)
Vậy \(A=-3\)
\(A=\left(\frac{3}{5}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)
\(A=\frac{5}{3}-\frac{3}{7}+9-2-\frac{5}{7}+\frac{2}{3}+\frac{8}{7}-\frac{4}{3}-10\)
\(A=\left(\frac{5}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{7}+\frac{5}{7}-\frac{8}{7}\right)+\left(9-2-10\right)\)
\(A=1+0-3\)
\(A=-2\)
Vậy \(A=-2\)
a) \(\frac{2}{5}+\left(-\frac{4}{3}\right)+\left(-\frac{1}{2}\right)\)
b) \(\frac{5}{8}-\left(-\frac{2}{5}-\frac{3}{10}\right)\)
c) \(\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\left(\frac{1}{4}+\frac{3}{8}\right)\right]\)
d) \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
e) \(\left(-\frac{1}{3}+\frac{5}{6}\right).11-7\)
f) \(\left(-\frac{5}{9}\right).\frac{3}{11}+\left(-\frac{13}{18}\right).\frac{3}{11}\)
thực hiện phép tính rồi rút gọn
A=\(\left(\frac{4}{9}+\frac{5}{6}\right):\left(\frac{7}{8}-\frac{7}{12}\right)\)
B=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
C=\(\left[3-\frac{1}{5}+\frac{3}{10}\right]:\left(2+\frac{1}{4}-\frac{3}{5}\right)\)
Tính nhanh :
A=\(21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)
B=\(\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)
C= \(\frac{-3}{5}+\frac{2}{7}+\frac{-3}{7}\cdot\frac{3}{5}+\frac{-3}{7}\)
\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)
\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)
\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)
\(A=\frac{154}{11}+\frac{8}{5}\)
\(\Rightarrow A=\frac{78}{5}\)
\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)
\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)
\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)
\(B=\frac{27}{9}+\frac{-72}{13}\)
\(B=3+\frac{-72}{13}\)
\(\Rightarrow B=\frac{-33}{13}\)
P/s: Hoq chắc :v
\(C=\frac{-3}{5}+\frac{2}{7}+\frac{-3}{7}.\frac{3}{5}+\frac{-3}{7}\)
\(C=\left(\frac{-3}{5}+\frac{3}{5}\right).\left(\frac{2}{7}+\frac{-3}{7}+\frac{-3}{7}\right)\)
\(C=0.\frac{-4}{7}\)
\(C=0\)