-Sửa đề: x,y nguyên.

\(x-\dfrac{1}{y}-\dfrac{4}{xy}=-1\left(x\ne0;y\ne0;x\ne-1\right)\)

\(\Rightarrow x-\dfrac{1}{y}-\dfrac{4}{xy}+1=0\)

\(\Rightarrow\dfrac{x^2y}{xy}-\dfrac{x}{xy}-\dfrac{4}{xy}+\dfrac{xy}{xy}=0\)

\(\Rightarrow x^2y-x-4+xy=0\)

\(\Rightarrow xy\left(x+1\right)=x+4\)

\(\Rightarrow y=\dfrac{x+4}{x\left(x+1\right)}\)

-Vì x,y nguyên: 

\(\Rightarrow\left(x+4\right)⋮\left[x\left(x+1\right)\right]\)

\(\Rightarrow\left(x+4\right)⋮x\) và \(\left(x+4\right)⋮\left(x+1\right)\)

\(\Rightarrow4⋮x\) và \(\left(x+1+3\right)⋮\left(x+1\right)\)

\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(3⋮\left(x+1\right)\)

\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(x+1\in\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(x\in\left\{0;-2;2;-4\right\}\)

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

*\(x=2\Rightarrow y=\dfrac{2+4}{2.\left(2+1\right)}=1\)

\(x=-2\Rightarrow y=\dfrac{-2+4}{-2.\left(-2+1\right)}=1\)

\(x=-4\Rightarrow y=\dfrac{-4+4}{-4.\left(-4+1\right)}=0\left(loại\right)\)

-Vậy các cặp số (x,y) là: \(\left(2,1\right);\left(-2,1\right)\)

\(x^2+4x+5=x^2+4x+4+1\)

\(=\left(x+2\right)^2+1\)

Ta có:

\(\left(x+2\right)^2\text{≡}0,1\left(mod3\right)\)

\(1\text{≡}1\left(mod3\right)\)

\(\Rightarrow\left(x+2\right)^2+1\text{≡}1,2\left(mod3\right)\)

\(\Rightarrow\left(x+2\right)^2+1\) không chia hết cho 3

\(\Rightarrow x^2+4x+5\) không chia hết cho 3

Ta có:\(\left|x-1\right|\ge0;\forall x\)

        \(\left|x+2\right|\ge0;\forall x\)

          \(\left|x-3\right|\ge0;\forall x\)

           \(\left|x+4\right|\ge0;\forall x\) ......

Cộng tất cả ta được:

\(\left|x-1\right|+\left|x+2\right|+\left|x-3\right|+\left|x+4\right|+...+\left|x-9\right|\ge0\)

\(\Rightarrow Min_T=0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x=1\\x=-2\\x=3\\x=-4.....\end{matrix}\right.\)

\(T=\left|9-x\right|+\left|x+8\right|+...+\left|3-x\right|+\left|x+2\right|+\left|x-1\right|\)

\(T\ge\left|9-x+x+8\right|+\left|7-x+x+6\right|+...+\left|3-x+x+2\right|+\left|x-1\right|\)

\(T\ge17+13+9+5+\left|x-1\right|\)

\(T\ge44+\left|x-1\right|\ge44\)

\(T_{min}=44\) khi \(x=1\)

A

A. (x-2)³ = x³ - 6x² +12x - 8 (hằng đẳng thức)

A

7. B

8. A

9. C

10. C

1B

8A

9C

10C

B A C C

3x3 – 6x2 + 3x = 3x(x2 - 2x + 1) = 3x(x - 1)2

Mik tặng coin thật mà, sao ko ai trl 

12.C

13.C

uầy nhìn huy hịu khỏng lò sợ tht

tăng 2coin nek mại dô:(

Toán hình à chị, e chịu r e có 6 điểm hình thôi .-.