Cho:
\(A=4+\frac{1}{7^6}+\frac{3}{7}+\frac{4}{7^2}+\frac{-441}{7^6}+\frac{27}{7^5}\)
\(B=\frac{147}{7}+4+\frac{35}{7^7}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\)
Hãy so sánh A với B
Cho C= \(4+\frac{1}{7^6}+\frac{3}{7}+\frac{4}{7^2}+\frac{-441}{7^6}+\frac{27}{7^5}\)
D=\(\frac{147}{7^3}+4+\frac{35}{7^7}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\)
So sánh C với D
So sánh C và D biết:
\(C=4+\frac{1}{7^6}+\frac{3}{7}+\frac{-441}{7^6}+\frac{27}{7^5}\)
\(D=\frac{147}{7^3}+4+\frac{35}{7}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\)
tính :
a) \(\frac{4}{9}:\frac{-1}{7}+6\frac{5}{9}:\frac{-1}{7}\)
b) \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
c) \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)
\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)
\(\frac{100}{-\frac{100}{41}}=-41\)
a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)
\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)
\(7:-\frac{1}{7}=-49\)
b. \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
\(\left(\frac{7}{3}+\frac{7}{2}\right):\left(-\frac{25}{6}+\frac{22}{7}\right)+\frac{15}{2}\)
\(\frac{35}{6}:-\frac{43}{42}+\frac{15}{2}\)
\(-\frac{245}{43}+\frac{15}{2}=\frac{155}{86}\)
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
\(B=\frac{1}{2}\)+\(\frac{-1}{5}\)+\(\frac{-5}{7}\)+\(\frac{1}{6}\)+\(\frac{-3}{35}\)+\(\frac{1}{3}\)+\(\frac{1}{41}\)
E=40+\(\frac{3}{8^{ }}\)+\(\frac{7}{8^2}\)+\(\frac{5}{8^3}\)+\(\frac{32}{8^5}\)
F=\(\frac{147}{7^3}\)+4+\(\frac{35}{7^7}\)+\(\frac{4}{7^2}\)+\(\frac{27}{7^5}\)+\(\frac{-9}{7}\)
đây là toán lớp 6 lên các bạn đừng thắc mắc
a: \(B=\left(-\dfrac{1}{5}-\dfrac{5}{7}+\dfrac{-3}{35}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)+\dfrac{1}{41}\)
\(=\dfrac{-7-25-3}{35}+\dfrac{3+2+1}{6}+\dfrac{1}{41}=\dfrac{42}{41}-1=\dfrac{1}{41}\)
Tính:
a) \(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)
b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)
c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)
\(C=\frac{2}{3}\)
Bạn Cô nàng Thiên Bình làm đúng hết òi =.=
a=7.[1/8+1/27-1/49]
------------------------
11.[1/8+1/27-1/49]
=7/11
cau b,c tuong tu nha h mk
a)\(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)
\(A=\frac{7.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}{11.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}\).
\(A=\frac{7}{11}\)
b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)
\(B=\frac{\frac{8}{9}.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}{4.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}\)
\(B=\frac{8}{9}:4=\frac{2}{9}\)
c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)\(C=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}\)
C=\(\frac{2}{3}\)
1. So sánh A và B:
\(A=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\); \(B=\frac{5}{7^3}+\frac{6}{7^2}+\frac{5}{7^4}+\frac{4}{7}+5\)
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,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)
