10x + 15 / 5x+1 =2x. (5x+ 1) + 13/ 5x + 1

= 1 + 2x + 13 / 5x + 1   để  A/frac/ 2x + 13/ 5x+1 nhận giá trị nguyên thì :

2x + 13 phải chia hết cho 5x +1 , ta có :

2x + 13 = 5x +1 

=> 2x + 5x = 13 +1

=> 7x =14

=> x= 2

Vậy x = 2 thì A có giá trị nguyên

\(A=\frac{10x+15}{5x+1}=\frac{2\left(5x+1\right)+13}{5x+1}=\frac{2\left(5x+1\right)}{5x+1}+\frac{13}{5x+1}\)

\(\Rightarrow5x+1\inƯ\left(13\right)=\left(-13;-1;1;13\right)\)

Ta có: \(5x+1=-13\Rightarrow x=-\frac{14}{5}\left(loại\right)\)

          \(5x+1=-1\Rightarrow x=-\frac{2}{5}\left(loại\right)\)

          \(5x+1=1\Rightarrow x=0\left(chọn\right)\)

          \(5x+1=13\Rightarrow x=\frac{12}{5}\left(loại\right)\)

Vậy x=0

2 dòng đầu không hiểu cho lắm

\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\left[\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\left[\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\frac{x^2-\left(x^2-10x+25\right)}{x\left(x-5\right)\left(x+5\right)}:\frac{10x-25}{x\left(x+5\right)}+\frac{x}{5-x}\)

\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{10x-25}+\frac{x}{5-x}\)

\(=\frac{1}{x-5}-\frac{x}{x-5}\)

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

Để P nguyên <=> x - 5 thuộc Ư(4) = {1;-1;2;-2;4;-4}

Ta có bảng:

Vậy....

\(ĐKXĐ:x\ne0;x\ne\pm5;x\ne\frac{5}{2}\)

\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)

ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)

b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)

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

\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)

a) Ta có: \(M=\dfrac{8x+1}{4x-5}=\dfrac{8x-10+11}{4x-5}=\dfrac{2\left(x-5\right)+11}{4x-5}=2+\dfrac{11}{4x-5}\)

Để M nhận giá trị nguyên thì \(2+\dfrac{11}{4x-5}\) nhận giá trị nguyên

\(\Rightarrow\dfrac{11}{4x-5}\) nhận giá trị nguyên

\(\Rightarrow11⋮4x-5\)

Vì \(x\in Z\) nên \(4x-5\in Z\)

\(\Rightarrow4x-5\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

\(\Rightarrow x\in\left\{1;\pm1,5;4\right\}\)

Vậy \(x\in\left\{1;4\right\}\) thỏa mãn \(x\in Z\).

b) Ta có: \(A=\dfrac{5}{4-x}\). ĐK: \(x\ne4\)

Nếu 4 - x < 0 thì x > 4 \(\Rightarrow A>0\)

       4 - x > 0 thì x < 4 \(\Rightarrow A< 0\)

Để A đạt GTLN thì 4 - x là số nguyên dương nhỏ nhất

\(\Rightarrow4-x=1\Rightarrow x=3\)

\(\Rightarrow A=\dfrac{5}{4-3}=5\)

Vậy MaxA = 5 tại x = 3

c) \(B=\dfrac{8-x}{x-3}\). ĐK: \(x\ne3\).

Ta có: \(B=\dfrac{8-x}{x-3}=\dfrac{-\left(x-8\right)}{x-3}=\dfrac{-\left(x-3\right)+5}{x-3}=\dfrac{5}{x-3}-1\)

Để B đạt giá trị nhỏ nhất thì \(\dfrac{5}{x-3}-1\) nhỏ nhất

\(\Rightarrow\dfrac{5}{x-3}\) nhỏ nhất

Nếu x - 3 > 0 thì x > 3 \(\Rightarrow\dfrac{5}{x-3}>0\) 

       x - 3 < 0 thì x < 3 \(\Rightarrow\dfrac{5}{x-3}< 0\)

Để \(\dfrac{5}{x-3}\) nhỏ nhất thì x - 3 là số nguyên âm lớn nhất

\(\Rightarrow x-3=-1\Rightarrow x=2\)

\(\Rightarrow B=\dfrac{8-2}{2-3}=-6\)

Vậy MaxB = -6 tại x = 2.

a) Để M nhận giá trị nguyên thì \(8x+1⋮4x-1\)

\(\Leftrightarrow8x-2+3⋮4x-1\)

mà \(8x-2⋮4x-1\)

nên \(3⋮4x-1\)

\(\Leftrightarrow4x-1\inƯ\left(3\right)\)

\(\Leftrightarrow4x-1\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow4x\in\left\{2;0;4;-2\right\}\)

\(\Leftrightarrow x\in\left\{\dfrac{1}{2};0;1;-\dfrac{1}{2}\right\}\)

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

Vậy: \(x\in\left\{0;1\right\}\)

\(a,A=\dfrac{x\left(x+2\right)+\left(2-x\right)\left(x-2\right)+12-10x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2x-4-x^2+2x+12-10x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-4x+8}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=-\dfrac{4}{x+2}\)

Vậy \(A=-\dfrac{4}{\left(x+2\right)}\)

a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

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

= \(\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)

=\(\frac{-3+x}{\left(x-2\right)\left(x-3\right)}.\frac{x+1}{1-x}\)

=\(\frac{1}{\left(x-2\right)}.\frac{x+1}{1-x}\)

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

b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)

\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)

\(\Leftrightarrow\frac{x-3}{x-2}>0\)

\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)

\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy ...