Tính:
\(\left(-2\frac{1}{5}\right)\times\left(\frac{-9}{11}\right)\times\left(-1\frac{1}{14}\right)\times\frac{2}{5}+\left(\frac{-4}{7}+\frac{6}{11}\right)\div\)\(\frac{-3}{5}\)
1) Rút gọn biểu thức M:
\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{2}{11}}\)
2) Tính nhanh:
\(A=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times....\times\left(1-\frac{1}{100}\right)\)
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
Tính nhanh nếu có thể......\(9\frac{2}{9}+\frac{2}{3}+7\frac{7}{9}\) \(\frac{5}{9}\times\frac{10}{11}+\frac{5}{9}\times\frac{14}{11}-\frac{5}{9}\times\frac{15}{11}\)
\(\frac{1}{6}\div\left(0,75+\frac{1}{6}\right)-50\%-\left(1\right)^2\)
\(9\frac{2}{9}+\frac{2}{3}+7\frac{7}{9}\)
\(=9+\frac{2}{9}+\frac{2}{3}+7+\frac{7}{9}\)
\(=\left(9+7\right)+\left(\frac{2}{9}+\frac{7}{9}\right)+\frac{2}{3}\)
\(=16+1+\frac{2}{3}\)
\(=17+\frac{2}{3}\)
\(=\frac{51}{3}+\frac{2}{3}\)
\(=\frac{53}{3}\)
\(\frac{5}{9}.\frac{10}{11}+\frac{5}{9}.\frac{14}{11}-\frac{5}{9}.\frac{15}{11}\)
\(=\frac{5}{9}\left(\frac{10}{11}+\frac{14}{11}-\frac{15}{11}\right)\)
\(=\frac{5}{9}.\frac{9}{11}\)
\(=\frac{5}{11}\)
\(\frac{1}{6}\div\left(0,75+\frac{1}{6}\right)-50\%-1^2\)
\(=\frac{1}{6}\div\left(\frac{9}{12}+\frac{2}{12}\right)-\frac{1}{2}-1\)
\(=\frac{1}{6}\div\frac{11}{12}-\frac{1}{2}-\frac{2}{2}\)
\(=\frac{1}{6}.\frac{12}{11}-\frac{3}{2}\)
\(=\frac{2}{11}-\frac{3}{2}\)
\(=\frac{4}{22}-\frac{33}{22}\)
\(=\frac{-29}{22}\)
Tính theo cách lớp 6 :
\(\left(a\right)-\frac{3}{7}\div\frac{11}{5}+-\frac{3}{7}\div\frac{11}{6}+\frac{9}{7}\)
\(\left(b\right)\frac{4}{13}+-\frac{11}{5}+\frac{9}{13}+\frac{6}{5}-75\%\)
\(\left(c\right)-\frac{19}{17}\times\frac{4}{7}+\frac{19}{17}\times-\frac{3}{7}+1\frac{2}{17}\)
Làm được bao nhiêu hay bấy nhiêu ! Làm hết 3 tick, làm 2 câu 2 tick, 1 câu 1 tick. Sai ko tính
a) = -3/7 . 5/11 + -3/7 . 6/11 + 9/7
= -3/7. ( 5/11 + 6/11 ) + 9/7
= -3/7. 1 + 9/7
= -3/7 + 9/7
= 6/7
b) = 4/13 + 9/13 + -11/5 + 6/5 - 3/4
= 13/13 + -5/5 - 3/4
= 1 + (-1) - 3/4
= 0 - 3/4
= -3/4
c) = -19/17. 4/7 + 19/17. -3/7 + 19/17
= 19/17. -4/7 + 19/17. -3/7 + 19/17.1
= 19/17.( -4/7 + -3/7 + 19/17
= 19/17. -7/7 + 19/17
= 19/17. (-1) + 19/17
= -19/17 + 19/17
= 0
tk mk nha,thanks
\(A=\left(1\frac{1}{6}\times\frac{6}{7}\times6:\frac{3}{5}\right):\left(4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{10}\right)\)
\(B=1\frac{13}{15}\times25\%\times3+\left(\frac{8}{15}-\frac{79}{60}\right):1\frac{23}{4}\)
\(C=\frac{123}{4567}\times\frac{1}{8}+\frac{123}{4567}\times\frac{1}{2}-\frac{123}{4567}\times\frac{13}{8}\)
\(D=\frac{10\frac{1}{3}\times\left(24\frac{1}{2}-15\frac{6}{7}\right)-\frac{12}{11}\times\left(\frac{10}{3}-1,75\right)}{\left(\frac{5}{9}-0,25\right)\times\frac{60}{11}+194\frac{8}{99}}\)
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 giá trị biểu thức
a, A=\(\frac{-1}{2}-\left[\frac{-3}{5}\right]+\left[\frac{-1}{9}\right]+\frac{1}{27}+\frac{7}{18}+\frac{4}{35}-\left[-\frac{2}{7}\right]\)
b, B=\(\frac{1}{3}-\frac{3}{4}-\left[\frac{-3}{5}-\frac{1}{57}+\frac{1}{36}+\frac{-1}{15}\right]-\frac{2}{9}\)
c, C=\(\left[-\frac{7}{15}\right]\times\frac{5}{8}\times\left[\frac{30}{-7}\right]\times\left[-16\right]\times\left[\frac{-1}{1000}\right]\)
d, D=\(\frac{1}{2}\times\frac{-11}{19}-50\%\times\left[-\frac{1}{19}\right]+\frac{10}{19}\times\frac{1111}{2222}\)
tính giá trị biểu thức chứ còn cái gì nữa
a, \(A=\frac{22}{27}\)
b,\(B=\frac{1}{57}\)
C,\(C=\frac{1}{50}\)
d, \(D=0\)
tính hợp lí nếu có thể
\(-\frac{1}{4}\times13\frac{9}{11}-0,25\times6\frac{2}{11}\)
\(B=\frac{-5}{6}\times\frac{4}{19}+\frac{-7}{12}\times\frac{4}{19}-\frac{40}{57}\)
\(\frac{3}{7}\times\frac{9}{26}-\frac{1}{14}\times\frac{1}{13}-\frac{1}{7}\)
\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
a) \(-\frac{1}{4}.13\frac{9}{11}-0,25.6\frac{2}{11}\)
\(=-\frac{1}{4}.13\frac{9}{11}-\frac{1}{4}.6\frac{2}{11}\)
\(=-\frac{1}{4}\left(13\frac{9}{11}+6\frac{2}{11}\right)\)
\(=-\frac{1}{4}.20\)
\(=-5\)
b) \(B=\frac{-5}{6}.\frac{4}{19}+\frac{-7}{12}.\frac{4}{19}-\frac{40}{57}\)
\(=\frac{4}{19}\left(\frac{-5}{6}+\frac{-7}{12}\right)-\frac{40}{57}\)
\(=\frac{4}{19}.\frac{-17}{12}-\frac{40}{57}\)
\(=\frac{-17}{57}-\frac{40}{57}\)
\(=-1\)
c) \(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}-\frac{1}{7}\)
\(=\frac{3}{7}.\frac{9}{26}-\frac{1}{2}.\frac{1}{7}.\frac{1}{13}-\frac{1}{7}\)
\(=\frac{1}{7}\left(3.\frac{9}{26}-\frac{1}{2}.\frac{1}{13}-1\right)\)
\(=\frac{1}{7}.0\)
\(=0\)
d) \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
\(=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)
\(=7:\left(-\frac{1}{7}\right)\)
\(=-49\)
\(A=-\frac{1}{4}x3\frac{9}{11}-0,25x6\frac{2}{11}\)
\(A=-0,25x3\frac{9}{11}-0,25x6\frac{2}{11}\)
\(A=-0,25x\left(\frac{9}{11}+\frac{2}{11}\right)\)
\(A=-0,25x1\)
\(A=-0,25\)
\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{40}{57}\)
\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{4}{19}x190\)
\(B=\frac{4}{19}x\left(-\frac{5}{6}-\frac{7}{12}-190\right)\)
\(B=\frac{4}{19}x\left(-\frac{2297}{12}\right)\)
\(B=\frac{-2297}{57}\)
\(C=\frac{3}{7}x\frac{9}{26}-\frac{1}{14}x\frac{1}{13}-\frac{1}{7}\)
\(C=\frac{1}{7}x3x\frac{9}{26}-\frac{1}{7}x\frac{1}{2}x\frac{1}{13}-\frac{1}{7}\)
\(C=\frac{1}{7}x\left(3x\frac{9}{26}-\frac{1}{2}x\frac{1}{13}-1\right)\)
\(C=\frac{1}{7}x0\)
\(C=\frac{1}{7}\)
\(D=\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
\(D=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)
\(D=7:\left(-\frac{1}{7}\right)\)
\(D=-49\)
\(4\times\left(\frac{1}{4}\right)^2+25\times\left[\left(\frac{3}{4}\right)^3\div\left(\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)
\(2^3+3\times\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2\div\frac{1}{2}\right]-8\)
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
1. Tính :
a.\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
b.\(\left(1-\frac{1}{7}\right)\times\left(1-\frac{2}{7}\right)\times\left(1-\frac{3}{7}\right)\times......\times\left(1-\frac{10}{7}\right)\)
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
Ta có \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
= \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
= \(\frac{2}{3}+\frac{1}{11}\)
= \(\frac{25}{33}\)