\(\left(2^2+2^1+2^2+2^3\right).2^0.2^1.2^2.2...">
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Tính hop ly

21.49-11.49+90.49+49.125.16

= 49.( 21 - 11 + 90 + 125 . 16 )

= 49 . ( 10 + 90 + 2000 )

= 49 . 2100

= 102900

(22+21+22+23).20.21.22.23

= ( 4 + 2 + 4 + 8 ) . 1 . 2 . 4 . 8

= 18 . 64

= 1152

Study well 

\(21.49-11.49+90.49+49.125.16\)

\(=49.\left(21-11+90+125.16\right)\)

\(=49.\left(10+90+2000\right)=49.2100\)

\(=102900\)

3 tháng 8 2019

(22+21+22+23).20.21.22.23

=(4+2+4+8).1.2.4.8

=18.1.2.4.8

=1152

1 3/8+1/8:(0,75-1/2)-25%.1/2

=11/8+1/8:(3/4-1/2)-1/4.1/2

=12/8:1/4-1/8

=6/1-1/8

=47/8

12 1/3-5/6:(24-23 5/7)

=37/3-5/6:(24-166/7)

=37/3-5/6:2/7

=37/3-35/2

=31/6

(-1/2)2-(-2)2-50

=1/4-4-1

=-19/4

3 tháng 8 2019

\(\left(2^2+2^1+2^2+2^3\right)×2^0×2^1×2^2×2^3\)

\(=\left(4+2+4+8\right)×1×2×4×8\)

\(=18×1×2×4×8\)

\(=1152\)

\(1\frac{3}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%×\frac{1}{2}\)

\(=\frac{11}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}×\frac{1}{2}\)

\(=\frac{11}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)

\(=\frac{11}{8}+\frac{1}{2}-\frac{1}{8}\)

\(=\frac{7}{4}\)

\(12\frac{1}{3}-\frac{5}{6}:\left(24-23\frac{5}{7}\right)\)

\(=\frac{37}{3}-\frac{5}{6}:\left(24-\frac{166}{7}\right)\)

\(=\frac{37}{3}-\frac{5}{6}:\frac{2}{7}\)

\(=\frac{37}{3}-\frac{35}{12}\)

\(=\frac{113}{12}\)

\(\left(\frac{-1}{2}\right)^2-\left(-2\right)^2-5^0\)

\(=\frac{1}{4}-4-1\)

\(=\frac{-19}{4}\)

13 tháng 6 2020

\(b,\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{100}{2}\)

\(=50\)

11 tháng 8 2016

c=1+2+4+8+64

=79

11 tháng 8 2016

\(C=2^0+2^1+2^2+2^3+2^0+2^1+2^2+2^3\)

\(=\left(2^0.2\right)+\left(2^1.2\right)+\left(2^2.2\right)+\left(2^3.2\right)\)

\(=2+4+8+16\)

\(=\left(2+8\right)+\left(4+16\right)\)

\(=10+20\)

\(=30\)

\(D=\left(2^0+2^1+2^2+2^3\right).2^0.2^1.2^2.2^3\)

\(=\left(1+2+4+8\right).1.2.4.8\)

\(=\left(8+2+4+1\right).1.2.4.8\)

\(=\left(10+4+1\right).1.2.4.8\)

\(=15.1.2.4.8\)

\(=\left(15.2\right).1.4.8\)

\(=30.1.4.8\)

\(=120.8\)

\(=960\)

14 tháng 2 2018

\(\left(1+2^2+2^3+....+2^{10}\right)\cdot\left(3^2\cdot2^4-12^2\right)\)

\(=\left(1+2^2+2^3+....+2^{10}\right)\cdot\left(3^2\cdot2^{2\cdot2}-12^2\right)\)

\(=\left(1+2^2+2^3+....+2^{10}\right)\cdot\left(3^2\cdot2^2\cdot2^2-12^2\right)\)

\(=\left(1+2^2+2^3+...2^{10}\right)\cdot\left[\left(3\cdot2\cdot2\right)^2-12^2\right]\)

\(=\left(1+2^2+2^3+....+2^{10}\right)\cdot\left[12^2-12^2\right]\)

\(=\left(1+2^2+2^3+...+2^{10}\cdot0\right)\)

\(=0\)

15 tháng 10 2016

Nguyễn Huy Tú cậu on ko giúp tớ với

 

15 tháng 10 2016

mik ko biết bài này Leona