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a) \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right].\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right).\left(-3\right)-25\)
\(=\left(-7\right)-120-25\)
\(=-152\)
b) \(\left(-2\right)^3.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=\left(-8\right).3-\left(1+8\right):9\)
\(=\left(-24\right)-9:9\)
\(=\left(-24\right)-1\)
\(=-25\)
Bài giải
a, \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right]\cdot\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right)\cdot\left(-3\right)-25\)
\(=-7-120-25\)
\(=-127-25\)
\(=-152\)
b, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)
\(=-8\cdot3-\left(1+8\right)\text{ : }9\)
\(=-24-9\text{ : }9\)
\(=-24-1\)
\(=-25\)
a)
Cách 1.
\(\begin{array}{l}\left( {4 + 32 + 6} \right) + \left( {10 - 36 - 6} \right)\\ = 4 + 32 + 6 + 10 - 36 - 6\\ = 52 + \left( { - 36} \right) + \left( { - 6} \right)\\ = 52 + \left( { - 42} \right) = 52 - 42 = 10\end{array}\)
Cách 2.
\(\begin{array}{l}\left( {4 + 32 + 6} \right) + \left( {10 - 36 - 6} \right)\\ = 4 + 32 + 6 + 10 - 36 - 6\\ = 36 + 6 + 10 + \left( { - 36} \right) + \left( { - 6} \right)\\ = 36 + \left( { - 36} \right) + 6 + \left( { - 6} \right) + 10\\ = 0 + 0 + 10 = 10\end{array}\)
b) \(\left( {77 + 22 - 65} \right) - \left( {67 + 12 - 75} \right)\)
\(\begin{array}{l} = 77 + 22 - 65 - 67 - 12 + 75\\ = 77 - 67 + 22 - 12 + 75 - 65\\ = 10 + 10 + 10 = 30\end{array}\)
c) \( - \left( { - 21 + 43 + 7} \right) - \left( {11 - 53 - 17} \right)\)
\(\begin{array}{l} = 21 - 43 - 7 - 11 + 53 + 17\\ = 21 - 11 + 53 - 43 + 17 - 7\\ = 10 + 10 + 10 = 30\end{array}\)
Đặt A bằng biểu thức trên
Ta có:
\(A=\left(1-\frac{4}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{2011}\right)\)
\(A=\frac{3}{7}.\left(\frac{7}{8}\right).\left(\frac{8}{9}\right)...\left(\frac{2009}{2010}\right).\left(\frac{2010}{2011}\right)\)
\(A=\frac{3}{7}.\left(\frac{\left(7.8...2009.2010\right)}{\left(8.9...2010.2011\right)}\right)\)
\(A=\frac{3}{7}.\frac{7}{2011}\)\(=\frac{3}{2011}\)
Ok nhé, bài này khá dễ !
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
-72(15-49) + 15 (-56 + 72)
= -72 . -34 + 15 . 16
= 2488 + 240
= 2728
-72(15-49)+15(-56+72) =-72.(-34)+15.16 =2448+240 = 2688 làm luôn :16.17.1.15625.1 =272.15625 =4250000
10 - { [ ( x : 3 + 17 ) : 10 + 3 : 24 ] : 10 } = 5
[ ( x : 3 + 17 ) : 10 + 3 : 24 ] : 10 = 10 - 5 = 5
( x : 3 + 17 ) : 10 + 3 : 24 = 5 x 10
( x : 3 + 17 ) : 10 + 48 = 50
( x : 3 + 17 ) : 10 = 50 - 48
( x : 3 + 17 ) : 10 = 2
x : 3 + 17 = 2 x 10
x : 3 + 17 = 20
x : 3 = 20 - 17 = 3
x = 3 x 3 = 9
a) [(2x+14) : 4 - 3] : 2 = 1
(2x+14) : 4 - 3 = 1/2
(2x+14) : 4 = 1/2 + 3
(2x+14) : 4 = 7/2
2x+14 = 7/2 . 1/4
2x = 7/8 - 1/4
2x = 5/8
x= 5/8.1/2
x= 5/16
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
\(a,-7-\left[\left(-19\right)+\left(21\right)\right].\left(-3\right)-\left[\left(32\right)+\left(-7\right)\right]\)
\(=-7-\left[21-19\right].\left(-3\right)-\left[32-7\right]\)
\(=-7-2.\left(-3\right)-25\)
\(=-7+6-25=-26\)
\(b,\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=4.3-9:9\)
\(=12-1=11\)