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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, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)
\(=-8\cdot3-\left(1+8\right)\text{ : }9\)
\(=-24-9\text{ : }9\)
\(=-24-1\)
\(=-25\)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
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
\(\left(\frac{2}{3}\right)^{21}\cdot\left(\frac{3}{2}\right)^{19}=\left(\frac{2}{3}\right)^{21}\cdot\left(\frac{2}{3}\right)^{-19}=\left(\frac{2}{3}\right)^2\)
\(\left(\frac{2}{5}\right)^3\cdot\left(\frac{5}{4}\right)^2=\frac{2^3}{5^3}\cdot\frac{5^2}{4^2}=\frac{4\cdot2}{5^2\cdot5}\cdot\frac{5^2}{4^2}=\frac{2}{20}=\frac{1}{10}\)
\(\left(\frac{5}{3}\right)^{27}\cdot\left(\frac{3}{5}\right)^{30}=\left(\frac{3}{5}\right)^{-27}\cdot\left(\frac{3}{5}\right)^{30}=\left(\frac{3}{5}\right)^3\)
câu cuối tương tự như câu 3
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
-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
a,\(\left(x-3\right).\left(2y+1\right)=7\)
Vì \(x;y\inℤ=>x-3;2y+1\inℤ\)
\(=>x-3;2y+1\inƯ\left(7\right)\)
Nên ta có bảng sau
| x-3 | 1 | 7 | -7 | -1 |
| 2y+1 | 7 | 1 | -1 | -7 |
| x | 4 | 10 | -4 | 2 |
| y | 3 | 0 | -1 | -4 |
Vậy ...
b,\(A=-126-\left(4^2-5\right)^2+870:29\)
\(=-126-\left(16-5\right)^2+30\)
\(=-126-11^2+30\)
\(=-247+30=-217\)
\(a,-7-\left[\left(-19\right)+\left(21\right)\right].\left(-3\right)-\left[\left(32\right)+\left(-7\right)\right]\)
\(=-7-\left[21-19\right].\left(-3\right)-\left[32-7\right]\)
\(=-7-2.\left(-3\right)-25\)
\(=-7+6-25=-26\)
\(b,\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=4.3-9:9\)
\(=12-1=11\)