\(\ge\forall\le\)Hình như cho Icons :>>

\(\forall\)là dấu : với mọi 

Ví dụ : \(\forall x\)thì \(x^2\ge0\)

Thế nhá

\(x^3+x^2-x-1\)

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

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

\(=\left(x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)\)

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

A = x³ + x² - x - 1

    = x²( x + 1 ) - ( x + 1 )

    = ( x + 1 ) ( x² - 1 )

    = ( x + 1 ) ( x - 1 ) ( x + 1 )

    = ( x + 1 )² ( x - 1 )

Chúc bạn học tốt nha!!!!!

\(25x^2-10xy^2+y^4=\left(5x-y^2\right)^2.\)

Ta có 25x²-10xy²+y⁴

=(5x-y²)² (cái này là hằng đẳng thức thứ 2 nha !!!!)

Xong rùi,nhớ

\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]\)

\(=\left(x^2-1\right)\left[x^4-2x^2+1-\left(x^4-x^2+1\right)\right]\)

\(=-3x^2\left(x^2-1\right)\)

   ( x² - 1 )³ - ( x⁴ + x² + 1 ) . ( x² - 1 )

= [ ( x² )³  - 3 . ( x² )² . 1 + 3 . x² . 1² - 1³ ] . [ ( x² )³ - 1³ ]

= ( x⁶ - 3x⁴ + 3x² - 1 ). ( x⁶ - 1 )

Mình không biết đề là gì nhưng mình nghĩ là phân tích đa thức thành nhân tử nên mình làm vậy, nếu đúng thì nhé.

I and Hung are team mates in our school’s soccer team, and that is the reason we became best friends. At first we did not know each other, but we quickly became close after just a few weeks. Hung is a good player, so he always helps me prace to improve my skills. In order to thank him, I become his instructor in some of the subjects in class. I am as tall and slim as Hung, and many people say that we look brothers. In fact, we are even closer than brothers. We can share almost everything, from feelings to clothes and hobbies. I always proud of our friend ship, and we will keep it this as long as we can.

Each summer vacation, I am taken to the countryside by my parents so as to relax after a hard working term. I really the life here. In rural areas, there are not as many houses and vehicles as in the city. As the result, the atmosphere is quite fresh and cool. Most people mainly travel by bicycle or on foot. The landscape is very poe and picturesque. There are many gardens, rivers here. We can grow vegetables or chickens, fish,... on our own. The local are very kind, generous and friendly. They are willing to give directions to strangers and help each other in difficult times. Whenever we are away, we can relievedly ask our neighbors to keep our houses and children . Another thing that we especially is that in the countryside, we are reconciled to nature so that the soul is always open and the body is healthy. Living in the countryside is fun and wonderful. In the future, I really want to live here.
 

ĐKXĐ : \(x\ne\pm3\)

a) \(A=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)

\(A=\left(\frac{-2x\left(3+x\right)}{\left(3-x\right)\left(3+x\right)}-\frac{\left(x+1\right)\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}+\frac{x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3}{x+3}-\frac{x-1}{x+3}\right)\)

\(A=\left(\frac{-2x^2-6x+x^2-2x-3+x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3-x+1}{x+3}\right)\)

\(A=\left(\frac{-8x-2}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{4}{x+3}\right)\)

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

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

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

b) \(\left|x-5\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}}\)

Mà ĐKXĐ x khác 3 => ta xét x = 7

\(A=\frac{4\cdot7+1}{2\cdot\left(7-3\right)}=\frac{29}{8}\)

c) Để A nguyên thì 4x + 1 ⋮ 2x - 3

<=> 4x - 6 + 7 ⋮ 2x - 3

<=> 2 ( 2x - 3 ) + 7 ⋮ 2x - 3

Mà 2 ( 2x - 3 ) ⋮ ( 2x - 3 ) => 7 ⋮ 2x - 3

=> 2x - 3 thuộc Ư(7) = { 1; -1; 7; -7 }

=> x thuộc { 2; 1; 5; -2 }

Vậy .....

a)   ĐKXĐ: \(x\ne\pm3\)

   \(A=\frac{2x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(x^2+1\right)}{x^2-9} : \frac{x+3-\left(x-1\right)}{x+3}\)

 \(A=\frac{2x^2-6x-x^2+2x+3-x^2-1}{x^2-9} : \frac{4}{x+3}\)

\(A=\frac{-4x+2}{x^2+9} : \frac{4}{x+3}\)

\(A=\frac{2\left(1-2x\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{4}=\frac{1-2x}{2x-6}\)

b)

  Có 2 trường hợp:

T.Hợp 1:

               \(x-5=2\Leftrightarrow x=7\)(thỏa mã ĐKXĐ)

thay vào A ta được: A=\(-\frac{13}{8}\)

T.Hợp 2:

          \(x-5=-2\Leftrightarrow x=3\)(Không thỏa mãn ĐKXĐ)

Vậy không tồn tại giá trị của A tại x=3

Vậy với x=7 thì A=-13/8

c)

      \(\frac{1-2x}{2x-6}=\frac{1-\left(2x-6\right)-6}{2x-6}=-1-\frac{5}{2x-6}\)

Do -1 nguyên, để A nguyên thì \(-\frac{5}{2x-6}\inℤ\)

Để \(-\frac{5}{2x-6}\inℤ\)thì \(2x-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Do 2x-6 chẵn, để x nguyên thì 2x-6 là 1 số chẵn .

Vậy không có giá trị nguyên nào của x để A nguyên

a,\(M=\left(\frac{4}{x-4}-\frac{4}{x+4}\right).\frac{x^2+8x+16}{32}\)

\(M=\left(\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\right).\frac{\left(x+4\right)^2}{32}\)

\(M=\frac{4x+16-4x+16}{\left(x+4\right)\left(x-4\right)}.\frac{\left(x+4\right)^2}{32}\)

\(M=\frac{32\left(x+4\right)^2}{32\left(x+4\right)\left(x-4\right)}=\frac{x+4}{x-4}\)

b,

Để M = \(\frac{1}{3}\)

\(\Rightarrow x-4=3x+12\)

\(\Rightarrow2x=16\Leftrightarrow x=8\)

\(c,\)\(\frac{x+4}{x-4}=\frac{x-4+8}{x-4}\)

\(\Rightarrow x-4\inƯ\left(8\right)=\left(1;-1;2;-2;4;-4;8;-8\right)\)

\(\Rightarrow x-4\in\left(5;3;6;2;8;0;12;-4\right)\)

Vậy để M thuộc Z thì x phải thỏa mãn các điều kiện trên .