\(\frac{x^2-3}{x^2-1}\in Z\)=>x²-3 chia hết cho x²-1

=>(x²-1)-2 chia hết cho x²-1

=>2 chia hết cho x²-1

=>x²-1\(\in\){-2;-1;1;2}

=>x²\(\in\){-1;0;2;3}

=>x\(\in\){0}

vậy x=0

ta có: \(\frac{x^2-3}{x^2-11}=\frac{x^2-11+8}{x^2-11}=1+\frac{8}{x^2-11}\)

Để \(\frac{x^2-3}{x^2-11}\inℤ\)

=> 8/x² -11 thuộc Z

=> 8 chia hết cho x^2 -11

=> x^2 - 11 thuộc Ư(8)={1;-1;2;-2;4;-4;8;-8}

...

rùi bn lập bảng xét giá trị hộ mk nha!

\(\frac{x^2-3}{x^2-11}\inℤ\Leftrightarrow x^2-3⋮x^2-11\)

\(\Rightarrow x^2-11+8⋮x^2-11\)

     \(x^2-11⋮x^2-11\)

\(\Rightarrow8⋮x^2-11\)

\(\Rightarrow x^2-11\inƯ\left(8\right)\)

\(\Rightarrow x^2-11\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow x^2\in\left\{10;12;9;13;7;15;3;19\right\}\) x là số nguyên

\(\Rightarrow x=3\)

+/ \(\hept{\begin{cases}x-1=-1\\y+2=-7\end{cases}}=>\hept{\begin{cases}x=0\\y=-9\end{cases}}\)

+/ \(\hept{\begin{cases}x-1=-7\\y+2=-1\end{cases}}=>\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)

ĐS: Các cặp số (x; y) thỏa mãn là {2; 5}; {8; -1}; {0; -9}; {-6; -3}

(x-1)(y+2)=1.7=7.1=(-1).(-7)=(-7).(-1)

+/ \(\hept{\begin{cases}x-1=1\\y+2=7\end{cases}}=>\hept{\begin{cases}x=2\\y=5\end{cases}}\)

+/ \(\hept{\begin{cases}x-1=7\\y+2=1\end{cases}}=>\hept{\begin{cases}x=8\\y=-1\end{cases}}\)

 a)  ĐKXĐ của phương trình : \(4x^2+4x+1\ne0\)\(\Rightarrow x\ne-\frac{1}{2}\)

b)  \(P=\frac{4x^3+8x^2-x-2}{4x^2+4x+1}\)

\(\Rightarrow P=\frac{\left(4x^3-x\right)+\left(8x^2-2\right)}{\left(2x+1\right)^2}\)

 \(\Rightarrow P=\frac{x\left(4x^2-1\right)+2\left(4x^2-1\right)}{\left(2x+1\right)^2}\)

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

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

\(\Rightarrow P\left(x\right)=4x^2+6x-6-\left(6x+3\right)=0\)

 \(\Rightarrow P\left(x\right)=4x^2-9=0\)\(\Rightarrow P\left(x\right)=x^2=\frac{9}{4}\)

\(\Rightarrow P\left(x\right)=x^2=\sqrt{\frac{9}{4}}\)\(\Rightarrow P\left(x\right)=\frac{3}{2}\)

câu c)  cx tương tự 

a, x khác -1/2

b, x=\(\frac{\sqrt{7}}{2}\)

thank

\(\text{Đk:}x\ne-\frac{1}{2}\Rightarrow P=\frac{4x^2\left(x+2\right)-\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(4x^2-1\right)\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(2x-1\right)\left(x+2\right)}{2x+1}\)

\(=\frac{2x^2+4x-x-2}{2x+1}=\frac{3}{2}\Rightarrow2x^2+3x-2=3x+\frac{3}{2}\Leftrightarrow2x^2-\frac{7}{2}=0......\)

\(P\text{ nguyên }\Rightarrow2x^2+3x-2⋮2x+1\Leftrightarrow2x^2+3x-2-\left(x+1\right)\left(2x+1\right)⋮2x+1\Leftrightarrow-3⋮2x+1....\)

a) \(P=\frac{4x^3+8x^2+x-2}{4x^2+4x+1}=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)

ĐKXĐ :\(\left(2x+1\right)^2\ne0=>2x+1\ne0=>x\ne-\frac{1}{2}\)

b) \(P=\frac{3}{2}\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{2x+1}=\frac{3}{2}\Leftrightarrow4x^2-2x+8x-4=6x+3\)

\(\Rightarrow4x^2=7=>x^2=\frac{7}{4}=>x=\pm\sqrt{\frac{7}{4}}\)

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

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

để P nguyền khi zà chỉ khi

\(1⋮2x+1\)

\(=>2x+1\inƯ\left(1\right)=\pm1\)

=>\(\orbr{\begin{cases}2x+1=1\\2x+1=-1\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

\(\frac{x^3-3}{x^3-1}=\frac{x^3-1-2}{x^3-1}=1-\frac{2}{x^3-1}\) là số nguyên

<=> x³ - 1 \(\in\) Ư(2) = {-2; -1; 1; 2}

<=> x³ \(\in\) {-1; 0; 2; 3}

Vì x là số nguyên nên x \(\in\) {-1; 0}