Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

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

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

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

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

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

Câu 1: C

Câu 2: =x(x-2)*(x+2)

Bài 1 : 

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

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

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

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

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

a. (2x² - 4x)\(\left(x-\dfrac{1}{2}\right)\)

= 2x³ - x² - 4x² + 2

= 2x³ - 5x² + 2

b. (x² - 2x + 1)(x - 1)

= (x - 1)²(x - 1)

= (x - 1)³

c. 3(y - x)(y² + xy + x²)

= 3(y³ - x³)

= 3y³ - 3x³

d. (x - 1)(x + 1)(x - 2)

= (x² - 1)(x - 2)

= x³ - 2x² - x + 2x

= x³ - 2x² + x 

= x³ - x² - x² + x

= x²(x - 1) - x(x - 1)

= (x² - x)(x - 1)

= x(x - 1)(x - 1)

= x(x - 1)²

Bài 1 : 

a, \(\left(2x^2-3x-1\right)\left(5x+2\right)=10x^3+4x^2-15x^2-6x-5x-2\)

\(=10x^3-11x^2-11x-2\)

b, sửa đề :  \(\left(-x^2+2x-3\right)\left(4x^2-2x+3\right)\)

\(=-4x^4+2x^3-3x^2+8x^3-4x^2+6x-12x^2+6x-9\)

\(=-4x^4+10x^3-19x^2+12x-9\)

Bài 2 : 

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

Thay x = 1 ; y = 1 ; z = -1 vào biểu thức trên ta được 

\(B=\left(1+1\right)\left(-2+1\right)+\left(1-1\right)\left(y-z\right)=2.\left(-1\right)=-2\)

Trả lời:

Bài 1: 

a, ( 2x² - 3x - 1 ) ( 5x + 2 ) 

= 10x³ + 4x² - 15x² - 6x - 5x - 2

= 10x³ - 11x² - 11x - 2

b, ( - x² + 2x - 3 ) ( 4x² - 2 + 3 )

= - 4x⁴ - 2x² + 3x² + 8x³ - 4x + 6x - 12x² + 6 - 9

= - 4x⁴ + 8x³ - 11x² + 2x - 3

Bài 2:

B = ( 2x + y ) ( 2z + y ) + ( x - y ) ( y - z ) 

Thay x = 1, y = 1, z = - 1 vào B, ta được:

B = ( 2.1 + 1 ) [ 2.( - 1 ) + 1 ] + ( 1 - 1 ) [ 1 -  ( - 1 )

= ( 2 + 1 ) ( - 2 + 1 ) + 0 . ( 1 + 1 )

= 3 . ( - 1 ) + 0

= - 3

Bài 2 :

a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)

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

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

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

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

\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)

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

c) Thay x = 4 ta có :

\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)

Vậy.........

\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)

\(\left(5x-2\right)\left(25x^2+10x+4\right)\)

\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)

\(=\left(5x\right)^3-2^3\)

\(=125x^3-8\)

Bài 1

a,\(4x^2y^3.\frac{2}{4}x^3y=2x^5y^4\)

Bài 1 

a)  (6x⁴y² - 3x³y³) : 3x³y² = 6x⁴y²  : 3x³y² - 3x³y³ : 3x³y² = 2x - y

b)  (2x - 1)(x² - x + 3) = 2x³ - 2x² + 6x - x² + x - 3 = 2x³ - 3x² + 7x - 3

Bài 2

1)     (x - 2)² - (x - 3)² = (x - 2 - x + 3)(x - 2 + x - 3) = 2x - 5>

2)     4x² - 4xy + 2y² + 1 = (4x² - 4xy + y²) + y² + 1 = (2x - y)² + y² + 1 > 0 

vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)