tìm các số nguyên x để mỗi phân số sau đây là số nguyên

Ngô Hải Nam ơi bn trả lời giúp mik ik

bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên

a: (2x+3)(y-4)=12

mà 2x+3 lẻ

nên (2x+3;y-4)∈{(1;12);(-1;-12);(3;4);(-3;-4)}

=>(2x;y)∈{(-2;16);(-4;-8);(0;8);(-6;0)}

=>(x;y)∈{(-1;16);(-2;-8);(0;8);(-3;0)}

b; x(2y+1)-4y=3

=>x(2y+1)-4y-2=1

=>x(2y+1)-2(2y+1)=1

=>(x-2)(2y+1)=1

=>(x-2;2y+1)∈{(1;1);(-1;-1)}

=>(x;2y)∈{(3;0);(1;-2)}

=>(x;y)∈{(3;0);(1;-1)}

c: xy+2x+y+11=0

=>x(y+2)+y+2+9=0

=>(x+1)(y+2)=-9

=>(x+1;y+2)∈{(1;-9);(-9;1);(-1;9);(9;-1);(3;-3);(-3;3)}

=>(x;y)∈{(0;-11);(-10;-1);(-2;7);(8;-3);(2;-5);(-4;1)}

\(Q=\frac{x^2-4}{x\left(x-1\right)}:\frac{x^2+2x}{x-1}=\frac{\left(x^2-4\right)\cdot\left(x-1\right)}{x\left(x-1\right)\cdot\left[x\left(x+2\right)\right]}=\frac{x^2-4}{x^2\left(x+2\right)}\)

1) \(\left|x\right|< 10\)

\(\Leftrightarrow-10< x< 10\)

2) \(\left|x\right|>11\)

\(\Leftrightarrow\left[{}\begin{matrix}x< -11\\x>11\end{matrix}\right.\)

3) \(\left|x\right|\ge2x\left(\forall x\ge0\right)\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}x\le-2x\\x\ge2x\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}3x\le0\\x\le0\end{matrix}\right.\)

\(\Leftrightarrow x=0\) \(\left(thỏa.đk:x\ge0\right)\)

4) \(\left|x\right|\le-3x\left(\forall x\le0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\left(-3x\right)\\x\le-3x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\le0\\4x\le0\end{matrix}\right.\)

\(\Leftrightarrow x\le0\) \(\left(thỏa.đk\right)\)

thuylinh1542004

Thái Thị Thùy Linh

1: =>15-2x=-11

=>2x=26

hay x=13

\(\left|x-1\right|+\left|x+5\right|=\left|x-1\right|+\left|-x-5\right|\)

\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|x-1-x-5\right|\)

\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|-6\right|=6\)

dấu "=" xảy ra khi \(\left(x-1\right).\left(x+5\right)\ge0\)

\(\Rightarrow-5\le x\le1\)

Vậy x={-5,-4,-3,-2,-1,0,1}

b) \(\hept{\begin{cases}\left(2x-y+3\right)^4\ge0\\\left|y+2\right|\ge0\end{cases}}\)

mà \(\left(2x-y+3\right)^4+\left|y+2\right|=0\)

dấu "=" xảy ra khi \(\hept{\begin{cases}\left(2x-y+3\right)^4=0\\\left|y+2\right|=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=-2\end{cases}}\)

vậy \(x=-\frac{5}{2},y=-2\)

$\left|x-1\right|+\left|x+5\right|=\left|x-1\right|+\left|-x-5\right|$

$\Rightarrow \left|x-1\right|+\left|x+5\right|\ge \left|x-1-x-5\right|$

$\Rightarrow \left|x-1\right|+\left|x+5\right|\ge \left|-6\right|=6$

dấu "=" xảy ra khi $\left(x-1\right).\left(x+5\right)\ge 0$

$\Rightarrow -5\le x\le 1$

Vậy x={-5,-4,-3,-2,-1,0,1}

b) \hept{(2�−�+3)4≥0∣�+2∣≥0\hept{(2x−y+3)4≥0∣y+2∣≥0​

mà (2�−�+3)4+∣�+2∣=0(2x−y+3)4+∣y+2∣=0

dấu "=" xảy ra khi \hept{(2�−�+3)4=0∣�+2∣=0\hept{(2x−y+3)4=0∣y+2∣=0​

⇒\hept{�=−52�=−2⇒\hept{

\(x(x-5)(x+5)-(x+2)(x^2-2x+4)\)

\(\Leftrightarrow x(x^2-25)-(x^3+8)\)

\(\Leftrightarrow x^3-25x-x^3-8\)

\(\Leftrightarrow-25x=11\Leftrightarrow x=-\frac{11}{25}\)

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)\)

\(\rightarrow\)\(x\left(x^2-25\right)-\left(x^3+8\right)\)

\(\rightarrow\)\(x^3-25x-x^3-8\)

\(\rightarrow\)\(-25x=11\Leftrightarrow x=-\frac{11}{25}\)

a) ta có: \(VT=\left|x-2\right|+\left|x+7,5\right|=\left|2-x\right|+\left|x+7,5\right|\le\left|2-x+x+7,5\right|=9=VP.\)

Muốn \(\left|x-2\right|+\left|x+7,5\right|=9\)thì  \(x=0\)

b) (Cái này mình không biết đúng hay không, nếu không thì các bạn ý kiến nha!)

+) Giả sử x = 0:  \(PT\Rightarrow2\left|0+3\right|+\left|2\cdot0\right|+5=6+5=11\)(đúng)

+) Giả sử x > 0:

\(PT\Leftrightarrow2\left(x+3\right)+2x+5=11\)

\(\Leftrightarrow2x+6+2x+5=11\)

\(\Leftrightarrow4x+11=11\)

\(\Leftrightarrow4x=0\Rightarrow x=0\)

+) Giả sử x < 0:

\(PT\Leftrightarrow-2\left(x+3\right)-2x-5=11\)

\(\Leftrightarrow-2x-6-2x-5=11\)

\(\Leftrightarrow-4x-11=11\)

\(\Leftrightarrow-4x=22\Rightarrow x=-\frac{11}{2}\)

Thử lại: \(2\left|-\frac{11}{2}+3\right|+\left|-\frac{2.11}{2}+5\right|=\frac{2.5}{2}+6=5+6=11\)(đúng)

Vậy x = 0 hoặc \(x=-\frac{11}{2}\)

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3