A nguyên

=>10x-15+6 chia hết cho 2x-3

=>\(2x-3\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{2;1;3;0\right\}\)

\(C=\dfrac{10-9}{2x-3}=\dfrac{1}{2x-3}\in Z\)

\(\Rightarrow\left(2x-3\right)\inƯ\left(1\right)=\left\{1;-1\right\}\)

Do \(x\in Z\)

\(\Rightarrow x\in\left\{1;2\right\}\)

a: Để A là số nguyên thì 10x-9⋮2x-3

=>10x-15+6⋮2x-3

=>6⋮2x-3

mà 2x-3 lẻ

nên 2x-3∈{1;-1;3;-3}

=>2x∈{4;2;6;0}

=>x∈{2;1;3;0}

b: Để B nguyên thì x-10⋮x-5

=>x-5-5⋮x-5

=>-5⋮x-5

=>x-5∈{1;-1;5;-5}

=>x∈{6;4;10;0}

\(A=\frac{3x-4}{2x-3}=\frac{2x-3+x-1}{2x-3}=1+\frac{x-1}{2x-3}\)

Để A có giá trị nguyên thì

\(x-1⋮2x-3\Leftrightarrow2x-2⋮2x-3\)

\(\Rightarrow2x-3-\left(2x-2\right)⋮2x-3\Rightarrow1⋮2x-3\)

\(\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

Có bạn nào làm được câu b không??

A=\(\frac{3x-4}{2x-3}=\frac{2x-3+x-1}{2x-3}=1+\frac{x-1}{2x-3}\)

Để A có giá trị nguyên thì:

x-1 \(⋮\)2x-3

=> 2x-2 \(⋮\)2x-3

=> 2x-3-(2x-2) \(⋮\)2x-3

=> 1 chia hết cho 2x-3

2x-3=1.              hoặc.   2x-3=-1

x=1.                                x=2

Để A nguyên thì \(x^2-4x-4⋮x-7\)

\(\Rightarrow x^2+3x-7x-21+17⋮x-7\)

\(\Rightarrow\left(x-7\right)\left(x+3\right)+17⋮x-7\)

Mà \(\left(x-7\right)\left(x+3\right)⋮x-7\)

\(\Rightarrow17⋮x-7\)

\(\Rightarrow x-7\in\left\{1;17;-1;-17\right\}\)

\(\Rightarrow x\in\left\{8;24;6;-10\right\}\)

\(\text{A=}\frac{x^2-4x-4}{x-7}\)

\(=\frac{x^2-4x-21+17}{x-7}\)

\(=\frac{x^2+3x-7x-21}{x-7}+\frac{17}{x-7}\)

\(=\frac{x\left(x+3\right)-7\left(x+3\right)}{x-7}+\frac{17}{x-7}\)

\(=\frac{\left(x-7\right)\left(x+3\right)}{x-7}+\frac{17}{x-7}\)

\(=\left(x+3\right)+\frac{17}{x-7}\)

Vì \(3\in Z\)

\(\Leftrightarrow x+3\in Z\)

\(\Rightarrow\text{A}\in Z\text{ khi }\frac{17}{x-7}\in Z\)

\(\Leftrightarrow\left(x-7\right)\inƯ\left(17\right)=\left\{1;-1;17;-17\right\}\)

\(\Leftrightarrow x=\left\{8;6;24;-10\right\}\)

Vậy với \(x=\left\{-10;6;8;24\right\}\)thì A có giá trị nguyên

Để A là số nguyê thì:

\(x^2-4x-4⋮x-7\)

\(x^2+3x-7x-21+17⋮x-7\)

\(\left(x-7\right)\left(x-3\right)+17⋮x-7\)

Mà \(\left(x-7\right)\left(x-3\right)⋮x-7\)nên

\(17⋮x-7\)

\(x-7\inƯ\left(17\right)=\left\{-1;1;-17;17\right\}\)

x={8;24;6;-10}

Anh thấy ảnh lỗi em ạ

Mik ko thấy ảnh đâu bn ơi

\(\text{Để }\frac{10x^2-7x-5}{2x-3}nguyên\Rightarrow\left(10x^2-7x-5\right)⋮\left(2x-3\right)\)

\(\text{Ta có }10x^2-7x-5=10x^2-7x-12+7=\left(2x-3\right)\left(5x+4\right)+7\)\(Mà\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)\Rightarrow7⋮\left(2x-3\right)\)

\(\Rightarrow\left(2x-3\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\text{Vậy x }\in\left\{-2;1;2;5\right\}\)