bạn có phải là Quỳnh học lớp 6 ngày mai chủ nhật đi học cô Giang toán ko

mình cũng ko bit làm cơ

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Ta có :\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)(1)

\(\Rightarrow3N=1+\frac{1}{3}+...+\frac{1}{3^8}\)(2)

Lấy (2) - (1) ta có :

\(\Rightarrow2N=1-\frac{1}{3^9}\)

\(\Rightarrow N=\frac{1-\frac{1}{3^9}}{2}\)

ta có: \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)

\(\Rightarrow\frac{1}{3}N=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{10}}\)

\(\Rightarrow\frac{1}{3}N-N=\frac{1}{3^{10}}-\frac{1}{3}\)

\(\frac{2}{3}N=\frac{1}{3^{10}}-\frac{1}{3}\)

\(N=\frac{\frac{1}{3^{10}}-\frac{1}{3}}{\frac{2}{3}}\)

CHÚC BN HỌC TỐT!!!

\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^9}\)

\(3N=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^8}\)

\(3N-N=1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^8}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.......+\frac{1}{3^9}\right)\)

\(2N=1+\frac{1}{3}+\frac{1}{3^2}+..........+\frac{1}{3^8}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.....-\frac{1}{3^9}\)

\(2N=1-\frac{1}{3^9}\)

\(N=\frac{1-\frac{1}{3^9}}{2}\)

\(A=47.36+64.47+15\)

\(A=47.\left(36+64\right)+15\)

\(A=47.100+15\)

\(A=4700+15\)

\(A=4715\)

\(B=27+35+65+73+75\)

\(B=\left(27+73\right)+\left(35+65\right)+75\)

\(B=100+100+75\)

\(B=275\)

\(C=37+37.15+84.37\)

\(C=37.\left(1+15+84\right)\)

\(C=37.100\)

\(C=3700\)

\(D=\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+\frac{1}{23.24}\)

\(D=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+\frac{1}{23}-\frac{1}{24}\)

\(D=\frac{1}{20}-\frac{1}{24}\)

\(D=\frac{24}{480}-\frac{20}{480}\)

\(D=\frac{4}{480}=\frac{1}{120}\)

\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(E=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(E=1-\frac{1}{50}\)

\(E=\frac{49}{50}\)

Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

Vì a là số nguyên tố > 3 nên a có dạng a = 3k + 1 hoặc a = 3k + 2 \(\left(k\inℕ\right)\)

-Nếu a = 3k + 1 thì \(\left(a-1\right)\cdot\left(a+4\right)=\left(3k+1-1\right)\left(3k+1+4\right)=3k\left(3k+5\right)\)

TH1: k là số chẵn thì \(k\left(3k+5\right)⋮2\Rightarrow3k\left(3k+5\right)⋮6\Rightarrow\left(a-1\right)\left(a+4\right)⋮6\)

TH2: k là số lẻ thì \(3k+5⋮2\Rightarrow k\left(3k+5\right)⋮2\Rightarrow3k\left(3k+5\right)⋮6\Rightarrow\left(a-1\right)\left(a+4\right)⋮6\)

-Nếu a = 3k + 2 thì \(\left(a-1\right)\left(a+4\right)=\left(3k+2-1\right)\left(3k+2+4\right)=\left(3k+1\right)\left(3k+6\right)\)

Chứng minh tương tự như trên ta cũng được \(\left(a-1\right)\left(a+4\right)⋮6\)

Giả sử a < b

Ta có:

a + b < b + b = 2b < a.b ( vì a > 2)

=> a + b < a.b ( đpcm)

mk ko k cho cậu đâu

Ờ kệ you t cx ko cần

\(2\frac{4}{5}.3\frac{1}{8}\)                                                                              \(1\frac{1}{5}:1\frac{4}{5}\)

= \(\frac{14}{5}.\frac{25}{8}\)                                                                          = \(\frac{6}{5}:\frac{9}{5}\)

= \(\frac{35}{4}\)                                                                                    = \(\frac{2}{3}\)

tk mk nha

\(2\frac{4}{5}\cdot3\frac{1}{8}\)

\(=\frac{14}{5}\cdot\frac{25}{8}\)

\(=\frac{35}{4}\)

\(1\frac{1}{5}:1\frac{4}{5}\)

\(=\frac{6}{5}:\frac{9}{5}\)

\(=\frac{2}{3}\)

\(2\frac{4}{5}\)x \(3\frac{1}{8}\)

\(=\frac{14}{5}\)x \(\frac{25}{8}\)

\(=\frac{14x25}{5x8}\)\(=\frac{35}{4}\)

\(1\frac{1}{5}:1\frac{4}{5}\)

\(=\frac{6}{5}:\frac{9}{5}\)

\(=\frac{6}{5}x\frac{5}{9}=\frac{6}{9}=\frac{2}{3}\)

Bài 1:

Bài 1: 1/3+1/9+1/27+1/81+1/243+1/729

Đặt:
A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân A với 3 ta có:
\(Ax3=3+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow Ax3-S=3-\frac{1}{243}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2\)
\(\Rightarrow A=\frac{1093}{729}\)

Bài 2Câu hỏi của nguyễn phạm mimi - Toán lớp 6 | Học trực tuyến

Bài 3

nguồn:Tính nhanh: 3/2 + 3/8 + 3/32 + 3/128 + 3/512 

Bài 4Câu hỏi của VŨ THỊ MAI CA - Toán lớp 4 - Học toán với OnlineMath

Bài 5Câu hỏi của Phạm Dora - Toán lớp 4 - Học toán với OnlineMath

Bài 6 Câu hỏi của huỳnh thị thu uyên - Toán lớp 5 - Học toán với OnlineMath