16 - 10 + 9 + 1 = ???
A. 10
B. 20
C. 30
Rút gọn
a) 20 - 12 . 20/ -30 - 30 . 10
b) 11^9 . 3^15 . 27/ 3^10 .11^11. 3^8
Em cố gắng gõ latex nha
a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{2}{3}\)
b: \(=\dfrac{11^9\cdot3^{18}}{3^{18}\cdot11^{11}}=\dfrac{1}{121}\)
Rút gọn
a) 20 - 12 . 20/ -30 - 30 . 10
b) 11^9 . 3^15 . 27/ 3^10 .11^11. 3^8
a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{20}{30}=\dfrac{2}{3}\)
b: \(=\dfrac{11^9\cdot3^{18}}{3^{10}\cdot11^{11}\cdot3^8}=\dfrac{1}{121}\)
So sanh
a ] 10 mu 20 va 19 mu 10
b ] [ -5 ] mu 30 va [ -3 ] mu 50
c ] 64 mu 8 va 16 mu 12
a) Ta có: \(10^{20}=\left(10^2\right)^{10}=100^{10}\)
Mà \(100^{10}>19^{10}\)
\(\Rightarrow10^{20}>19^{10}\)
b) Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
Mà: \(125^{10}< 243^{10}\)
\(\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
c) Ta có: \(64^8=\left(2^6\right)^8=2^{48}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
Mà: \(2^{48}=2^{48}\)
\(\Rightarrow64^8=16^{12}\)
a) 1020và 1910
Ta có: 1020= (102)10 và 1910
= 10010 và 1910
Vì 10010>1910 => 1020>1910
b) (-5)30 và (-3)50
Ta có:
(-5)30= [(-5)3]10=(-125)10 và (-3)50=[(-3)5]10=(-243)10
Vì -12510>-24310 Nên (-5)30>(-3)50
c) 648 và 1612
= (43)8và (42)12
= 424 và 424
=> 648 = 1612
a ] Ta có : 10\(^{20}\) = 10\(^{^{ }2.10}\) = [ 10\(^2\) ]\(^{10}\) = 100\(^{10}\)
Vì 100\(^{10}\) > 19\(^{10}\) Nên => 10\(^{20}\) > 19\(^{10}\)
b ] Ta có : [ -5 ]\(^{30}\) = [ -5 ]\(^{3.10}\) = [ -5\(^3\) ]\(^{10}\) = [ -125 ]\(^{10}\)
[ -3 ]\(^{50}\) = [ -3 ]\(^{5.10}\) = [ -3\(^5\) ]\(^{10}\) = [ -243 ]\(^{10}\)
Vì [ -125 ]\(^{10}\) < [ -243 ]\(^{10}\) Nên => [ -5 ]\(^{30}\) < [ -3 ]\(^{50}\)
c ] Ta có : 64\(^8\) = 64\(^{2.4}\) = [ 64\(^2\) ]\(^4\) = 4196\(^4\)
16\(^{12}\) = 16\(^{2.6}\) = [ 16\(^2\) ]\(^6\) = 4096\(^4\)
Vì 4196\(^4\) > 4096\(^4\) Nên => 64\(^8\) > 16\(^{12}\)
Tìm x trong phép tính sau: 16 -x = 30 *
1 điểm
A = 46
B = -46
C =14
D = -14
Tìm x trong phép tính sau: x + 30 = 20 *
1 điểm
A = 10
B = -10
C = 50
D = -50
Chọn khẳng định đúng trong các khẳng định sau *
1 điểm
A.- 5 là số nguyên dương
B. -5 > 3
C.- 5 < 3
a) 20\16 - 3\4 b) 30\45 - 2\5
c) 10\12 -3\4 d)12\9 -1\4
a: =5/4-3/4=2/4=1/2
b: =30/45-18/45=12/45=4/15
c: =10/12-9/12=1/12
d: =4/3-1/4=16/12-3/12=13/12
a,2/3x-1/5x=-7/10
b,(1/10-1).(1/11-1).(1/12-1)....(1/2021-1)
c, 1/12+1/20+1/30+1/42+1/50+1/72+1/90
tính dùm mình nhé ,mình đang vội
a)
\(\dfrac{2}{3}x-\dfrac{1}{5}x=-\dfrac{7}{10}\\ \left(\dfrac{2}{3}-\dfrac{1}{5}\right)x=-\dfrac{7}{10}\\ \dfrac{7}{15}x=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}:\dfrac{7}{15}\\ x=-\dfrac{15}{10}\\ x=-\dfrac{3}{2}.\)
b)
\(\left(\dfrac{1}{10}-1\right).\left(\dfrac{1}{11}-1\right).\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{2021}-1\right)\\ =\dfrac{-9}{10}.\dfrac{-10}{11}.\dfrac{-11}{12}...\dfrac{-2020}{2021}\\ =\dfrac{9}{2021}.\)
c)
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\\ =\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{3}-\dfrac{1}{10}\\ =\dfrac{7}{30}.\)
(câu c em ghi nhầm đề 1 chút nhé, chỗ đó là 56 chứ không phải 50)
Tính nhanh:
a,1/4+2/5+6/8+9/15+8/1
b,1/2+2/4+3/6+4/8+5/10+6/12+7/14+8/16+9/18+10/20
c,1/10+4/20+9/30+16/40+25/50+36/60+49/70+64/80+81/90
Tính nhanh:
a,1/4+2/5+6/8+9/15+8/1
b,1/2+2/4+3/6+4/8+5/10+6/12+7/14+8/16+9/18+10/20
c,1/10+4/20+9/30+16/40+25/50+36/60+49/70+64/80+81/90
bài 2: tính nhanh :
a) 3/8 + 7/12 + 10/16 + 10/24 b) 4/6 + 7/13 + 17/9 + 19/13 + 1/9 + 14/6
c) 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56
a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)
\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)
\(=1+1\)
\(=2\)
b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=3+2+2\)
\(=7\)
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)