x/3 = -12/9

=> x/3 = -4/3

=> x = -4

vậy_

1.Ta có: \(\frac{x}{3}=-\frac{12}{9}\)

=> \(\frac{3x}{9}=-\frac{12}{9}\)

=> 3x = -12

=> x = -12 : 3

=> x = -4

\(\frac{4}{5}x-\frac{8}{5}=-\frac{1}{2}\)

=> \(\frac{4}{5}x=-\frac{1}{2}+\frac{8}{5}\)

=> \(\frac{4}{5}x=\frac{11}{10}\)

=> \(x=\frac{11}{10}:\frac{4}{5}\)

=> \(x=\frac{11}{8}\)

bài 1: \(\frac{x}{3}=\frac{-12}{9}\)=> 9x=-36

=> x=-4

vậy x=-4

\(\frac{4}{5}x-\frac{8}{5}=\frac{-1}{2}\)=> \(\frac{4}{5}x=\frac{-1}{2}+\frac{8}{5}\)

=> \(\frac{4}{5}x=\frac{-5}{10}+\frac{16}{10}\)=\(\frac{11}{10}\)=> \(x=\frac{11}{10}:\frac{4}{5}\)=\(\frac{11}{10}.\frac{5}{4}\)=\(\frac{11}{8}\)

vậy x=\(\frac{11}{8}\)

\(\frac{1}{5}.\left|x\right|-1\frac{2}{5}=\frac{2}{5}\)=> \(\frac{1}{5}.\left|x\right|-\frac{7}{5}=\frac{2}{5}\)

=> \(\frac{1}{5}.\left|x\right|=\frac{2}{5}+\frac{7}{5}=\frac{9}{5}\)=> |x| =\(\frac{9}{5}:\frac{1}{5}\)=9

=> x=9 hoặc x=-9

vậy x=9 hoặc x=-9

\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)

\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)

\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)

\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=3-\left(1-\frac{1}{8}\right)\)

\(A=3-\frac{5}{8}\)

\(A=\frac{19}{8}\)

a, Để phân số đạt giá trị nguyễn 

\(\Rightarrow x+1⋮x-2\)

\(\Rightarrow x-2+3⋮x-2\)

mà \(x-2⋮x-2\Rightarrow3⋮x-2\)

\(\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow x\in\left\{3;5\pm1\right\}\)

b,Tương tự :

\(2x-1⋮x+5\)

\(\Rightarrow2x+10-11⋮x+5\)

\(2\left(x+5\right)-11⋮x+5\)

mà \(2\left(x+5\right)⋮x+5\Rightarrow11⋮x+5\)

\(\Rightarrow x+5\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

\(x\in\left\{-4;\pm6;-16\right\}\)

a, Để \(A\in Z\)\(\Leftrightarrow x+1⋮x-2\)\

Ta có:              \(\hept{\begin{cases}x-2⋮x-2\\x+1⋮x-2^{ }_{ }\end{cases}}\) \(\Rightarrow\) \(x-2-\left(x+1\right)⋮x-2\)

\(\Rightarrow-3⋮x-2\)mà \(x\in Z\Rightarrow x-2\in Z\Rightarrow x-2\inƯ\left(-3\right)\)\(\in\left(1;-1,3;-3\right)\)

\(x\in\left(3;1;5;-1\right)\)Vậy: \(x\in\left(1;3;5;-1\right)\)thì \(A\in Z\)

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

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

Số 0 nha bn !

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{13}{12}\le x\le\frac{1}{24}-\left(-\frac{5}{24}\right)\)

\(\Rightarrow\frac{6}{12}-\frac{13}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow x\in\left\{-7;-6;-5;...;0;1;2;3\right\}\)

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}-\frac{3}{4}\le x\le\frac{1}{24}-\frac{1}{8}+\frac{1}{3}\)

\(\Rightarrow\frac{6-4-9}{12}\le x\le\frac{1-3+8}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{6}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow-7\le x\le3\)

\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3;....;2;3\right\}\)