m = 5 

n = -1

mình nhầm câu trên

\(=>\dfrac{2m}{10}+\dfrac{1}{10}=-\dfrac{1}{n}\)

\(=>\dfrac{2m+1}{10}=-\dfrac{1}{n}\)
\(=>n\left(2m+1\right)=\left(-10\right)\)
\(=>\left[{}\begin{matrix}n=1=>m=-\dfrac{11}{2}\left(loại\right)\\n=\left(-1\right)=>m=\dfrac{9}{2}\left(loại\right)\\n=10=>m=\left(-1\right)\left(tm\right)\\n=\left(-10\right)=>m=0\left(tm\right)\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}n=2=>m=-3\left(tm\right)\\n=-2=>m=2\left(tm\right)\\n=5=>m=-\dfrac{3}{2}\left(loại\right)\\n=\left(-5\right)=>m=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)

\(=>\)Các cặp (m,n) thỏa mãn là: (-1,10)(0,-10)(-3,2)(2,-2)
 

\(\dfrac{m}{5}+\dfrac{1}{10}=\dfrac{-1}{n}\left(n\ne0\right)\)

\(\Rightarrow\dfrac{2mn}{10n}+\dfrac{n}{10n}=\dfrac{-10}{10n}\)

\(\Rightarrow2mn+n=-10\)

\(\Rightarrow n\left(2m+1\right)=-10\)

\(\Rightarrow n=\dfrac{-10}{2m+1}\)

-Vì m,n ∈ Z.

\(\Rightarrow-10⋮\left(2m+1\right)\)

\(\Rightarrow2m+1\inƯ\left(10\right)\)

\(\Rightarrow2m+1\in\left\{1;2;5;10;-1;-2;-5;-10\right\}\)

\(\Rightarrow m\in\left\{0;2;-1;-3\right\}\)

\(m=0\Rightarrow n=\dfrac{-10}{2.0+1}=-10\)

\(m=2\Rightarrow n=\dfrac{-10}{2.2+1}=-2\)

\(m=-1\Rightarrow n=\dfrac{-10}{2.\left(-1\right)+1}=10\)

\(m=-3\Rightarrow n=\dfrac{-10}{2.\left(-3\right)+1}=2\)

-Vậy các cặp số (m,n) là (0,-10) ; (2,-2) ; (-1,10) ; (-3,2).

 

á đù m cũng dùng nhá

mà sao m có đề câu này thế

a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)

\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)

b. -Để M thuộc Z thì:

\(\left(x^2+x-2\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)

\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)

\(\Rightarrow4⋮\left(x+3\right)\)

\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)

c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)

\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)

câu 1

\(\dfrac{m}{2}\).\(\dfrac{2}{n}\)=\(\dfrac{1}{2}\)

\(\dfrac{m}{2}\).\(\dfrac{2}{n}\)=\(\dfrac{4}{8}\)

\(\dfrac{4}{8}\)=\(\dfrac{2.m}{2.n}\)

\(\dfrac{4}{8}\)=\(\dfrac{1.m}{1.n}\)

\(\dfrac{4}{8}\)=\(\dfrac{m}{n}\)=\(\dfrac{1}{2}\)

câu2

câu2

a/ta có;n+1/n-2

=n-2+3/n-2

để a là số ngyên thì n-2+3 phải chia hết cho n-2

xét n-2+3 có n-2 chia hết cho n-2 nên suy ra 3 cũng phải chia hết cho n-2

vậy n-2 là Ư(3)=1;-1;3;-3

nếu n-2=-1thì n=-1+2 ;n=1

nếu n-2=1 thì n=1+2;n=3

nếu n-2=-3 thì n=-3+2=-1(ko đúng với điều kiện đề bài cho)

nếu n-2=3 thì n= 3+2=5

còn phần b cậu thấy cái nao nhỏ nhất thì chọn

a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3

\(\Rightarrow\) n - 2 \(\in\) Ư(3)

\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}

n \(\in\){5; -1; 3; 2}

c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{3}-\dfrac{1}{30}\)

\(=\dfrac{10}{30}-\dfrac{1}{30}\)

\(=\dfrac{9}{30}\)

=\(\dfrac{3}{10}\)

d)\(\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right).\left(1-\dfrac{1}{6}\right).......\left(1-\dfrac{1}{29}\right).\left(1-\dfrac{1}{30}\right)\)\(=\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}....\dfrac{28}{29}\)

\(=\dfrac{3.4.5...28}{4.5.6...29}\)

\(=\dfrac{3}{29}\)

Bài 1 :

Sửa đề :

Tìm \(n\in Z\) để những phân số sau đồng thời có giá trị nguyên

\(\dfrac{-12n}{n};\dfrac{15}{n-2};\dfrac{8}{n+1}\)

Làm

Ta có :

\(\dfrac{-12n}{n}=-12\)

\(\Leftrightarrow\) Với mọi \(n\) thì \(\dfrac{-12n}{n}\) đều có giá trị nguyên \(\left(1\right)\)

Để \(\dfrac{15}{n-2}\in Z\) \(\Leftrightarrow n-2\inƯ\left(15\right)=\left\{\pm1;\pm15;\pm3;\pm5\right\}\)

\(\Leftrightarrow n\in\left\{-13;\pm3;\pm1;5;7;17\right\}\left(1\right)\)

Để \(\dfrac{8}{n+1}\in Z\Leftrightarrow n+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow n\in\left\{-9;-5;\pm3;-2;0;1;7\right\}\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow n\in\left\{\pm3;1;7\right\}\)