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b) 73.9+ 32.74- 45.539
=343.9+9.2401-45.539
=9.(343+2401)-45.539
=9.2744-24355
=24696-24355
=341
a) 132- (116-(132-1282))
=132- ( 116-(132-16384))
=132-(116+16252)
=132-16368
=16236
125(28+72)-25(3^2.4+64)
=125.100-25(9.4+64)
=125.100-25.(36+64)
=125.100-25.100
=12500-2500
=10000
\(a.132-\left[116-4^2\right]\) \(c.120:\left\{300:\left[150-\left(2.125-8.25\right)\right]\right\}\)
= \(132-\left[116-16\right]\) = \(120:\left\{300:\left[150-\left(250-200\right)\right]\right\}\)
= \(132-100\) = \(120:\left\{300:\left[150-50\right]\right\}\)
= \(32\) = \(120:\left\{300:100\right\}\)
= \(120:3\)
= \(40\)
132 - [ 116 - ( 132 - 128 )2 ] { 46 - [( 16 + 71 , 4 ) : 15 ]} -2 Ý cuối cùng mình làm luôn cứ thế mà
= 132 - [ 116 - 16 ] = { 46 - [ 87,4 : 15] } - 2 chép.
=132 - 100 = { 46 - 5,8266 } - 2 = 120 : { 300 : [ 150 - ( 210 - 200) ]}
= 32 = 40,1734 - 2 = 120 : { 300 : [ 150 - 10 ]}
=38,1734m = 120 : { 300 : 140 }
= 120 : 2,1428
= 56,0014
SAI THÌ THÔI NHA !
\(\left[116-\left(13-5^2\right)\right]=\left[116-\left(13-25\right)\right]=116-13+25=128\)
\(\frac{7^{25}}{7^{21}\times46+7^{21}\times3}=\frac{7^{25}}{7^{21}\times\left(46+3\right)}=\frac{7^4}{50}\)
\(57+212+43+288=\left(57+43\right)+\left(212+288\right)=100+500=600\)
\(28\times69+4\times18\times7+2\times14\times13=28\times69+18\times28+28\times13=28\times\left(69+18+13\right)=28\times100=2800\)
Bài 1:
a) Ta có: \(\frac{2^8\cdot4\cdot13+2^7\cdot8\cdot65}{2^9\cdot39}\)
\(=\frac{2^8\cdot4\cdot13+2^8\cdot4\cdot13\cdot5}{2^9\cdot39}\)
\(=\frac{2^{10}\cdot13\left(1+5\right)}{2^9\cdot13\cdot3}=\frac{6}{3}=2\)
b) Đặt \(A=4+2^2+2^3+2^4+...+2^{20}\)
Ta có: \(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow2A=2^3+2^3+2^4+...+2^{21}\)
Ta có: \(2A-A=2^3+2^{21}-2^2-2^2=8+2^{21}-8=2^{21}\)
hay \(A=2^{21}\)
Vậy: \(4+2^2+2^3+2^4+...+2^{20}=2^{21}\)
\(5^8:25^2=5^8:5^4\)
=54=625
\(4^9:64^2=4^9:\left(2^2\right)^6\)
=49:46
=43=64
58 : 252 = 58 : (52)2
= 58 : 54
= 54
49 : 642 = 49 : (43)2
= 49 : 46
= 43
\(=116:\left(100-81\right)=116:19=\dfrac{116}{19}\)