a) \(x^4-y^4\)

\(=\left(x^2\right)^2-\left(y^2\right)^2\)

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

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

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)

\(=\left(5x-5y\right)\left(x+y\right)\)

\(=5\left(x-y\right)\left(x+y\right)\)

d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)

\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)

\(=\left(5x-y\right)\left(x-5y\right)\)

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

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

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

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

f) \(x^3+27\)

\(=x^3+3^3\)

\(=\left(x+3\right)\left(x^2-3x+9\right)\)

g) \(27x^3-0,001\)

\(=\left(3x\right)^3-\left(0,1\right)^3\)

\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

THAM KHẢO NHÉ BẠN, MÌNH KO VT ĐC TRONG NÀY NÊN PHẢI RA WORD VÍT R COPY VÀO Á:

$7x^2-13x+6=0\iff 7x^2-7x-6x+6=0$

$\iff 7x\left(x-1\right)-6\left(x-1\right)=0\iff \left(7x-6\right)\left(x-1\right)=0$

$\iff \{\begin{array}{c}x-1=0\\ 7x-6=0\end{array}$ $\iff \{\begin{array}{c}x=1\\ 7x=6\end{array}$ $\iff \{\begin{array}{c}x=1\\ x=\frac{6}{7}\end{array}$ vậy $x=1;x=\frac{6}{7}$

`#040911`

`7x^2 - 13x + 6`

`= 7x^2 - 7x - 6x + 6`

`= (7x^2 - 7x) - (6x - 6)`

`= 7x(x - 1) - 6(x - 1)`

`= (7x - 6)(x - 1)`

Ta có: 

\(C=2x^2-4x+6\)

\(C=2\cdot\left(x^2-2x+3\right)\)

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

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

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

Mà: \(2\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow C=2\left(x-1\right)^2+4\ge4>0\forall x\)

Vậy tất cả các số thực đều thỏa mãn:

\(\Rightarrow x\in R\)

`C = 2x^2 - 4x + 6`

`2C = 4x^2 - 8x + 12`

`2C = ( 2x )^2 - 2 . 2x . 2 + 2^2 + 12 - 2^2`

`2C = ( 2x - 2 )^2 + 8`

Vì ` ( 2x - 2 )^2 >= 0 AAx` nên : 

`( 2x - 2 )^2 + 8 >= 8 > 0 AAx`

Hay `2C > 0 AAx` . Vì `2C > 0 AAx => C > 0 AAx` .

Vậy `C > 0 AAx` ( đpcm ) .

a) \(4x^2-1\)

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

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

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(9x^2-\dfrac{1}{4}\)

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

\(=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)

d) \(\left(x-y\right)^2-4\)

\(=\left(x-y\right)^2-2^2\)

\(=\left(x-y-2\right)\left(x-y+2\right)\)

e) \(9-\left(x-y\right)^2\)

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

\(=\left(3+x-y\right)\left(3-x+y\right)\)

f) \(\left(x^2+4\right)^2-16x^2\)

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

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

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

Bài 2 ạ

  đây ạ

\(\dfrac{x+2}{x-2}=\dfrac{x-2+4}{x-2}=\dfrac{x-2}{x-2}+\dfrac{4}{x-2}=1-\dfrac{4}{x-2}\) 

Nguyên khi:

\(\dfrac{4}{x-2}\) nguyên 

\(\Rightarrow4\) ⋮ \(x-2\)

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

\(\Rightarrow x\in\left\{3;1;4;0;6;-2\right\}\)

Vậy: ... 

Dạ mik cảm ơn 

Bài 2:

a)

ĐK: \(x\ne\pm2\)

\(A=\left(\dfrac{x+2}{x^2-4}-\dfrac{x-2}{x^2-4}\right):\dfrac{4}{\left(x+2\right)^2}\\ =\dfrac{x+2-x+2}{x^2-4}:\dfrac{4}{\left(x+2\right)\left(x+2\right)}\\ =\dfrac{4\left(x+2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right).4}\\ =\dfrac{x+2}{x-2}\)

b)

Thế \(x=-\dfrac{1}{2}\) vào A được:

\(A=\dfrac{x+2}{x-2}=\dfrac{-\dfrac{1}{2}+2}{-\dfrac{1}{2}-2}\\ =\dfrac{\dfrac{3}{2}}{-\dfrac{5}{2}}=\dfrac{3}{5}.\dfrac{-2}{2}\\ =\dfrac{3}{5}.-1=-\dfrac{3}{5}\)

c)

Để A nhận giá trị nguyên thì \(\left(x+2\right)⋮\left(x-2\right)\)

\(\Leftrightarrow\left(x-2+4\right)⋮\left(x-2\right)\\ \Rightarrow4⋮\left(x-2\right)\\ \Rightarrow\left(x-2\right)\in\left\{\pm1;\pm2;\pm4\right\}\\ \Rightarrow x\in\left\{3;1;4;0;6;-2\right\}\)

Bài thứ 2 ạ  

B  bài 1 ạ

Phần tự luận:

Bài 1:

a)

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

b)

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

Bài 2:

ĐK: \(x\ne\pm2\)

\(A=\left(\dfrac{x+2}{x^2-4}-\dfrac{x-2}{x^2-4}\right):\dfrac{4}{\left(x+2\right)^2}\\ =\dfrac{x+2-x+2}{x^2-4}:\dfrac{4}{\left(x+2\right)\left(x+2\right)}\\ =\dfrac{4\left(x+2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right).4}\\ =\dfrac{x+2}{x-2}\)

a) x² - 6x + 9

= x² - 2.x.3 + 3²

= (x - 3)²

b) x² - 6

= x² - (√6)²

= (x - √6)(x + √6)

c) 1 - 27x³

= 1³ - (3x)³

= (1 - 3x)(1 + 3x + 9x²)

d) x³ + 1/x³

= x³ + (1/x)³

= (x + 1/x)(x² - 1 + 1/x²)

e) -x³ + 9x² - 27x + 27

= -(x³ - 3.x².3 + 3.x.3² - 3³)

= -(x - 3)³

a)

\(x^2-6x+9=x^2-2.x.3+3^2\\ =\left(x-3\right)^2\)

b)

\(x^2-9\) mới đúng đề chứ 

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

Hoặc với `x^2-6` luôn nha:

\(x^2-6\\ =x^2-\left(\sqrt{6}\right)^2\\ =\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)

c)

\(1-27x^3\\ =1-\left(3x\right)^3\\ =\left(1-3x\right)\left(1+3x+9x^2\right)\)

d)

\(x^3+\dfrac{1}{x^3}\\ =x^3+\left(\dfrac{1}{x}\right)^3\\ =\left(x+\dfrac{1}{x}\right)\left(x^2-1+\dfrac{1}{x^2}\right)\)

e)

\(-x^2+9x^2-27x+27\\ =-\left(x^2-9x^2+27x-27\right)\\ =-\left(x^2-3.3x^2+3.3^2.x-3^3\right)\\ =-\left(x-3\right)^3\\ =-\left(x-3\right)\left(x-3\right)\left(x-3\right)\)