Câu 8( Mình không viết đè nữa nha)

a)   2-1/1.2 + 3-2/2.3 + 4-3/3.4 +…..+ 100-99/99.100

=  1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…..+ 1/99 – 1/100

=  1 – 1/100 < 1

=   99/100 < 1

    Vậy A< 1

Bài 1 :
a) =) \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)= \(1-\frac{1}{101}=\frac{100}{101}\)
b) =) \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=) \(\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)( theo phần a)
Bài 2 :
-Gọi d là UCLN \(\left(2n+1;3n+2\right)\)( d \(\in N\)* )
(=) \(2n+1⋮d\left(=\right)3.\left(2n+1\right)⋮d\)
(=) \(6n+3⋮d\)
và \(3n+2⋮d\left(=\right)2.\left(3n+2\right)⋮d\)
(=) \(6n+4⋮d\)
(=) \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
(=) \(6n+4-6n-3⋮d\)
(=) \(1⋮d\left(=\right)d\in UC\left(1\right)\)(=) d = { 1;-1}
Vì d là UCLN\(\left(2n+1;3n+2\right)\)(=) \(d=1\)(=) \(\frac{2n+1}{3n+2}\)là phân số tối giản ( đpcm )
Bài 3 :
-Để A \(\in Z\)(=) \(n+2⋮n-5\)
Vì \(n-5⋮n-5\)
(=) \(\left(n+2\right)-\left(n-5\right)⋮n-5\)
(=) \(n+2-n+5⋮n-5\)
(=) \(7⋮n-5\)(=) \(n-5\in UC\left(7\right)\)= { 1;-1;7;-7}
(=) n = { 6;4;12;-2}
Vậy n = {6;4;12;-2} thì A \(\in Z\)
Bài 4:
A = \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= \(10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)= \(10101.\left(\frac{2}{222222}+\frac{5}{222222}\right)\)
= \(10101.\frac{7}{222222}\)( không cần rút gọn \(\frac{7}{222222}\))
= \(\frac{7}{22}\)

các bạn giúp mình với mình đang cần đáp án gấp

1) a.Ta có \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)}{n-4}+\frac{21}{n-4}=3+\frac{21}{n-4}\)

Vì \(3\inℤ\Rightarrow\frac{21}{n-4}\inℤ\Rightarrow21⋮n-4\Rightarrow n-4\inƯ\left(21\right)\)

=> \(n-4\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)

=> \(n\in\left\{5;3;8;1;11;-3;25;-17\right\}\)

b) Ta có B = \(\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=\frac{3\left(2n-1\right)+8}{2n-1}=3+\frac{8}{2n-1}\)

Vì \(3\inℤ\Rightarrow\frac{8}{2n-1}\inℤ\Rightarrow2n-1\inƯ\left(8\right)\Rightarrow2n-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)(1)

lại có với mọi n nguyên => 2n \(⋮\)2 => 2n - 1 không chia hết cho 2 (2)

Kết hợp (1) ; (2) => \(2n-1\in\left\{1;-1\right\}\Rightarrow n\in\left\{1;0\right\}\)

2) Ta có : \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

=> \(\frac{20+xy}{4x}=\frac{1}{8}\)

=> 4x = 8(20 + xy)

=> x = 2(20 + xy)

=> x = 40 + 2xy

=> x - 2xy = 40

=> x(1 - 2y) = 40

Nhận thấy : với mọi y nguyên => 1 - 2y là số không chia hết cho 2 (1)

mà x(1 - 2y) = 40

=> 1 - 2y \(\inƯ\left(40\right)\)(2)

Kết hợp (1) (2) => \(1-2y\in\left\{1;5;-1;-5\right\}\)

Nếu 1 - 2y = 1 => x = 40

=> y = 0 ; x = 40

Nếu 1 - 2y = 5 => x = 8

=> y = -2 ; x = 8 

Nếu 1 - 2y = -1 => x = -40

=> y = 1 ; y = - 40

Nếu 1 - 2y = -5 => x = -8

=> y = 3 ; x =-8

Vậy các cặp (x;y) thỏa mãn là : (40 ; 0) ; (8; - 2) ; (-40 ; 1) ; (-8 ; 3)

4) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{-4}{3}}=\frac{-\frac{5}{60}}{\frac{2}{5}}=-\frac{5}{60}:\frac{2}{5}=-\frac{5}{24}\)

b) \(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}\)

\(=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)

c) \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

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Các bạn ơi giải giúp mink vs mink đg cần gấp

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LẠM DỤNG QUÁ NHIỀU

1. a) Để \(A=\frac{3n+5}{n+1}\)là phân số thì \(n+1\ne0\Leftrightarrow n\ne-1\)

Vậy ...

b) Để A là ps thì \(3n+5⋮n+1\)

Ta có: \(3n+5=3\left(n+1\right)+2\)

Vì \(3\left(n+1\right)⋮n+1\)nên để \(3n+5⋮n+1\)thì \(2⋮n+1\Leftrightarrow n+1\varepsilonƯ\left(2\right)\)

Bạn tự tìm n nha rồi kết luận

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).