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\(B=81.\left[\frac{3\left(12-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}\right)}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{79.2.1001001}{79.9.1001001}\)
\(B=81.\left[3.\frac{6}{5}\right].\frac{2}{9}\)
\(B=\frac{9.9.3.6.2}{5.9}\)
\(B=\frac{9.3.6.2}{5}\)
\(B=\frac{324}{5}\)
Tick cho minh nha Quang Hai Duong tick minh may man ca nam
a) 23.15 -[115-(12-5)2 ]
= 23.15 -[115-36]
= 8.15 -79
= 120-79
=41
b)5.[(85 - 35 : 7) :8 + 90 ] - 50
=5 .[80:8+90]-50
=5.100-50
=500-50
=450
c){[261 - ( 36-31)3.2 ]-9}.1001
={[261 - 125.2 ]-9}.1001
={[261 -250 ]-9}.1001
={11-9}.1001
=2.1001
=2002
d)3.102 - [1200 - ( 42 - 2.3)3]
=3.100-[1200 -(16-6)3 ]
=300-[1200-1000]
=300-200
=100
\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
Vậy \(A=-2\)
\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)
\(B=81.\frac{18}{5}.\frac{2}{9}\)
\(B=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
Chúc bạn học tốt ~ ( mỏi tay qué >_< )
P/s : nhìn thì khủng thật ! :v
\(B=81.\left[\frac{\left[12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}\right]}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158}{711}\)
\(B=81.\left(\frac{3}{1}:\frac{5}{6}\right).\frac{158}{711}\)
\(B=81.\frac{18}{5}.\frac{158}{711}\)
\(B=\frac{1458}{5}.\frac{158}{711}=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
A,
26 x 108 - 26 x 12
32-28 + 24 - 20 + 16- 12 + 8 -4
26x (108 - 12 )
4 + 4 + 4 +4
26 x 96
4x4
=156
a) = 8 . 15 - [115- 72 ]
= 8 . 15 - [115- 49 ]
= 120 - 56
= 64
b) = 100 : {250 : [450 - (2.125 - 8.25)]}
= 100 : {250 : [450 - (250 - 200 )]}
= 100 : {250 : [450 - 50]}
= 100 : {250 :400}
= 100 : 0,265
= 0,00265
=
\(a,2^3.15-\left[115-\left(12-5\right)^2\right]\)
\(=8.15-\left[115-\left(12-5\right)^2\right]\)
\(=120-\left[115-7^2\right]\)
\(=120-\left[115-49\right]\)
\(=120-66\)
\(=54\)
\(b,100:\left\{250:\left[450-\left(2.5^3-2^3.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(2.125-8.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(250-200\right)\right]\right\}\)
\(=100:\left\{250:\left[450-50\right]\right\}\)
\(=100:\left\{250:400\right\}\)
\(=100:0,625\)
\(=160\)
\(2^3.15-\left[115-\left(12-5\right)^2\right]=8.15-\left[115-7^2\right]=120-\left[115-49\right]=120-66=54\)
\(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}=100:\left\{250:\left[450-\left(4.125-4.25\right)\right]\right\}=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)
\(=100:\left[250:\left(450-400\right)\right]=100:\left(250:50\right)=100:5=20\)