ta có

x+y+y+z+z+x=\(\frac{13}{12}\)

2(x+y+z)=\(\frac{13}{12}\)

=>x+y+z=\(\frac{13}{24}\)

z=(x+y+z)-(x+y)

y=y+z-z

x=x+Y-y

Bài 3:

a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)

2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)

2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)

3A = \(1-\frac{1}{2^6}\)

=> 3A < 1 

=> A < \(\frac{1}{3}\)(đpcm)

b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)

4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)       (1)

Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)

4B = \(3-\frac{1}{3^{99}}\)

=> 4B < 3

=> B < \(\frac{3}{4}\)   (2)

Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)

bài 1:

5n+7 chia hết cho 3n+2

=> [3(5n+7) - 5(3n + 2)] chia hết cho 3n+2

=> (15n + 21 - 15n - 10) chia hết cho 3n+2

=> 11 chia hết cho 3n + 2

=> 3n + 2 thuộc Ư(11) = {1;-1;11;-11}

Ta có bảng:

Vậy n = {-1;3}

Bài 2:

1, chữ số tận cùng

a, Xét 7¹⁹⁹⁹

Ta có: 7¹⁹⁹⁹ = 7¹⁹⁹⁶.7³ = (7⁴)⁴⁹⁹.343 = (...1)⁴⁹⁹.343 = (....1).343 = ....3 (1)

Vậy số 57¹⁹⁹⁹ có tận cùng là 3

b, Xét 3¹⁹⁹⁹

Ta có: 3¹⁹⁹⁹ = 3¹⁹⁹⁶.3³ = (3⁴)⁴⁹⁹.27 = (...1)⁴⁹⁹.27 = (...1) . 27 = ....7  (2)

Vậy số 93¹⁹⁹⁹ có chữ số tận cùng là 7

2, 

Từ (1) và (2) suy ra A = 999993¹⁹⁹⁹ + 555557¹⁹⁹⁹ = ...7 + ...3 = ....0

Vì A có chữ số tận cùng là 0 nên A chia hết cho 5. 

\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{41.45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{8}{45}\)

\(\Rightarrow x=\frac{37}{45}+\frac{8}{45}\)

\(\Rightarrow x=1\)

Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

a, \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)

\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{7}-\frac{x+1}{9}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

b, \(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)

\(\Leftrightarrow\left(\frac{x+4}{96}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+1}{99}+1\right)\)

\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}=\frac{x+100}{98}+\frac{x+100}{99}\)

\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}-\frac{x+100}{98}-\frac{x+100}{99}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

a) x + 1/5 + x + 1/7 = x + 1/9

<=> 1/5x + 1/5 + 1/7x + 1/7 = 1/9x + 1/9

<=> (1/5x + 1/7x) + (1/5 + 1/7) = 1/9x + 1/9

<=> 12/35x + 12/35 = 1/9x + 1/9

<=> 12/35x + 12/35 - 1/9x = 1/9 

<=> 73/315x + 12/35 = 1/9

<=> 73/315x = 1/9 - 12/35

<=> 73/315x = -73/315

<=> x = 73/315 : -73/315 = -1

=> x = -1

b) làm tương tự

1+1=3 :)))

a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)

\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)

\(=\frac{1}{100}\)

b) Tương tự như câu a

Ra kết quả phần b là \(\frac{101}{2}\) đúng hông?

câu a ra kết quả là 1/100 nhé

      \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow\)\(x+329=0\)   (vì  1/327 + 1/326 + 1/325 + 1/324 + 1/5  khác  0  )

\(\Leftrightarrow\)\(x=-329\)

Bài 1 : 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

\(\Rightarrow\)\(x+329=0\)

\(\Rightarrow\)\(x=-329\)

Vậy \(x=-329\)

Bài 1: 

a ) = 12/21

b ) = 50

k cho mik nha

Các bn giải cụ thể ra giúp mk đc k? c. ơn các bn