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Nếu lượng sữa có  3,1% lượng bơ suy ra lượng sữa là:

100% - 3,1% = 96,9% ( chai )

Lượng sữa trong 1 chai là:

9,3 : 100 x 96,9 = 9,0117 ( g )

Đ/s: ......

~ Hok T ~

\(N=5+5^2+5^3+...+5^{49}+5^{50}\)

\(5N=5^2+5^3+5^4+...+5^{50}+5^{51}\)

\(5N-N=\left(5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(5+5^2+5^3+...+5^{49}+5^{50}\right)\)7

\(4N=5^2+5^3+5^4+...+5^{50}+5^{51}-5-5^2-5^3-...-5^{49}-5^{50}\)

\(4N=5^{51}-5\)

\(N=\frac{5^{51}-5}{4}\)

\(20\%x-x+\frac{1}{5}=\frac{3}{4}\)

\(\rightarrow x\left(\frac{1}{5}-1\right)=\frac{3}{4}-\frac{1}{5}\)

\(\rightarrow x\frac{-4}{5}=\frac{11}{20}\)

\(\rightarrow x=\frac{11}{20}:\frac{-4}{5}\)

\(\rightarrow x=\frac{-11}{16}\)

b) \(\frac{-5}{7}\).\(\frac{2}{11}\)--\(\frac{-5}{7}\).\(\frac{9}{11}\)+ \(1\frac{5}{7}\)

= \(\frac{-5}{7}\). (\(\frac{2}{11}\)--\(\frac{9}{11}\)) + \(1\frac{5}{7}\)

= \(\frac{-5}{7}\).  1 + \(1\frac{5}{7}\)

=\(\frac{-5}{7}\)+\(\frac{10}{7}\)

=\(\frac{5}{7}\)

c)\(11\frac{3}{13}\)-- (\(2\frac{4}{7}\)+\(5\frac{3}{13}\))

=\(11\frac{3}{13}\)-- \(2\frac{4}{7}\)-- \(5\frac{3}{13}\)

=(\(11\frac{3}{13}\)-- \(5\frac{3}{13}\)) -- \(2\frac{4}{7}\)

=6 -- \(\frac{18}{7}\)

=\(\frac{42}{7}\)--\(\frac{18}{7}\)

=\(\frac{24}{7}\)

\(A=1+3+3^2+3^3+...+3^{2019}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\)

\(=4+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)

\(=4\left(1+3^2+...+3^{2018}\right)⋮4\)

\(A=1+3+3^2+...+3^{2019}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{2020}\)

\(\Rightarrow3A-A=3^{2020}-1\)

\(\Rightarrow A=\frac{3^{2020}-1}{2}\)

Ta có \(3^{2020}\equiv1\left(mod4\right)\)

\(\Rightarrow\frac{3^{2020}-1}{2}\equiv0\left(mod4\right)\)

Suy ra A chia hết cho 4

a, x=2

b,x=16

c, x=3

d, x=3

còn lại thì chịu mới lớp  4 thôi

a) 2/5 = 2/5

b) 3/8 = 6/16

c) 1/9 = 3/27

d) 4/3 = 8/6

e) Lớp 7 r quên

f) -8/-2 = -8/-2

ĐÁp án :

Bạn ơi

Đề bài thiếu nhé

Học tốt !!!!!!!!!!!!!!!!!!!!!!!

Ta có A = 1 + 3 + 3² + 3³ + .... + 3²⁰¹⁸ + 3²⁰¹⁹

= (1 + 3) + 3²(1 + 3) + ... + 3²⁰¹⁸(1 + 3)

= (1 + 3)(1 + 3² + .... + 3²⁰¹⁸)

= 4(1 + 3² + .... + 3²⁰¹⁸) 

=> A \(⋮\)4 hay A : 4 dư 0