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: \(x\in\left\{0;1\right\}\)

Câu 1) ngộ thế

a: A>0

=>\(x^2-3x>0\)

=>x(x-3)>0

TH1: \(\left\{{}\begin{matrix}x>0\\x-3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>0\\x>3\end{matrix}\right.\)

=>x>3

TH2: \(\left\{{}\begin{matrix}x< 0\\x-3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\x< 3\end{matrix}\right.\)

=>x<0

d: Để D<0 thì \(x^2+\dfrac{5}{2}x< 0\)

=>\(x\left(x+\dfrac{5}{2}\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x>0\\x+\dfrac{5}{2}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>0\\x< -\dfrac{5}{2}\end{matrix}\right.\)

=>Loại

Th2: \(\left\{{}\begin{matrix}x< 0\\x+\dfrac{5}{2}>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\x>-\dfrac{5}{2}\end{matrix}\right.\)

=>\(-\dfrac{5}{2}< x< 0\)

e: ĐKXĐ: x<>2

Để E<0 thì \(\dfrac{x-3}{x-2}< 0\)

TH1: \(\left\{{}\begin{matrix}x-3>=0\\x-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\x< 2\end{matrix}\right.\)

=>Loại

TH2: \(\left\{{}\begin{matrix}x-3< =0\\x-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\x>2\end{matrix}\right.\)

=>2<x<=3

g: Để G<0 thì \(\left(2x-1\right)\left(3-2x\right)< 0\)

=>\(\left(2x-1\right)\left(2x-3\right)>0\)

TH1: \(\left\{{}\begin{matrix}2x-1>0\\2x-3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>\dfrac{3}{2}\end{matrix}\right.\)

=>\(x>\dfrac{3}{2}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\2x-3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< \dfrac{3}{2}\end{matrix}\right.\)

=>\(x< \dfrac{1}{2}\)

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 ...