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a) ( 2600 + 6400 ) - 3 . x = 1200
9000 - 3 . x = 1200
3 . x = 9000 - 1200
3 . x = 7800
x = 7800 : 3
x = 2600
b) [ ( 6 . x - 72 ) : 2 - 84 ] . 28 = 5628
[ ( 6 . x - 72 ) : 2 - 84 ] = 5628 : 28
[ ( 6 . x - 72 ) : 2 - 84 ] = 201
( 6 . x - 72 ) : 2 = 201 + 84
( 6 . x - 72 ) : 2 = 285
( 6 . x - 72 ) = 285 . 2
( 6 . x - 72 ) = 570
6 . x = 570 + 72
6 . x = 642
x = 642 : 6
x = 107
c) 2 . x - 138 = 23 . 32
2 . x - 138 = 8 . 9
2 . x - 138 = 72
2 . x = 72 + 138
2 . x = 210
x = 210 : 2
x = 105
d, 42 . x = 39 . 42 - 37 . 42
42 . x = 42 . ( 39 - 37 )
42 . x = 42 . 2
\(\Rightarrow\) x = 2
\(B=\frac{2^3.17-8.14}{96}=\frac{8.17-8.14}{8.12}=\frac{8\left(17-14\right)}{8.12}=\frac{3}{12}=\frac{1}{4}\)
C = ( 39 . 42 - 37 . 42 ) : 42
= 42 ( 39 - 37 ) : 42
= 42 . 2 : 42
= 2
D = 2448 : [ 119 - ( 62 - 9 ) ] : 20180
= 2448 : [ 119 - 27 ] : 1
= 2448 : 172 : 1
= \(\frac{612}{23}\)
Bài 1:
a) A=1+22+24+.................+2100
2A=(1+22+24+.................+2100)
2A=2+23+...+2101
2A-A=(2+23+...+2101)-(1+22+24+.................+2100)
A=2101-1
b)bạn tự làm
c) C=-1/90-1/72-1/50-1/42-1/30-1/20-1/12-1/6-1/2
\(=-\left(\frac{1}{90}+\frac{1}{72}+...+\frac{1}{2}\right)\)
\(=-\left(\frac{1}{10.9}+\frac{1}{9.8}+...+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{10}-1\right)\)
\(=-\left(-\frac{9}{10}\right)=\frac{9}{10}\)
Bài 2:
cứ tính lần lượt là ra
a. 3 x 52-- 16:22
= 3 \(\times\)25 - 16 : 4
= 75 - 4
= 71
b. (39x42 --37x42) : 42
= 42 \(\times\)(39 - 37) : 42
= 42 \(\times\)2 : 42
= 84 : 42
= 2
1448 : [119 - (23-6)]
= 1448 : (119 - 17)
= 1448 : 102
= \(\frac{724}{51}\)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
b, (39. 42 - 37 . 42 ) : 42
= [ 42 ( 39 - 37 ) ] : 42
= 2 . 42 : 42
= 2
a, A = 3 . 42 - 81 : 32
A = 3 . 16 - 81 : 9
A = 48 - 9
A = 39
b, B = 1999 + [ 100 : ( 8 - 6 )2 - 22 . 3 ]
B = 1999 + [ 100 : 22 - 4 . 3 ]
B = 1999 + [ 100 : 4 - 12 ]
B = 1999 + [ 25 - 12 ]
B = 1999 + 13
B = 2012
\(3^6:3^2+2^3.2^2=3^{6-2}+2^{3+2}\)
\(=3^2+2^5=9+32=41\)
\(\left(39.42-37.42\right):42=\left[42.\left(39-37\right)\right]:42\)
\(=\left(42.2\right):42=84:42=2\)
\(2x-138=2^3.3^2\Leftrightarrow2x-138=8.9\)
\(2x-138=72\Leftrightarrow2x=210\Leftrightarrow x=105\)
\(231-\left(x-6\right)=1339:13\Leftrightarrow231-\left(x-6\right)=103\)
\(\Leftrightarrow x-6=128\Leftrightarrow x=128+6\Leftrightarrow x=134\)
a, 36: 32+ 23.22
= 34+25
= 81+32
= 113
b, (39. 42- 37. 42) : 42
= (39-37).42/42
= 2 . ( 42/42)
= 2*1
=2
Bài 2: Tìm số tự nhiên x biết
a, 2.x - 138=23.32 b, 231- (x - 6) = 1339:13
2*x-138 = 8*9 231 - (x-6) =103
2*x-138 = 72 x - 6 = 128
2*x = 210 x = 134
x =105
a) 36:32+23.22
= 36-2+23+2
= 34+25= 81 + 32 = 113 (nhấc điện thoại lên nào bạn ơi)
b) (39.42-37.42):42
= [(39 - 37).42)] : 42
= (2.42):42
= 84 : 42 = 2
a, 36:33+23.22
=36-2+23+2
=34+25
=81+32
=113