Bài 2:

\(A=-2x^2+3x-5\)

\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)

\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)

\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)

Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)

Bài 1:

a)x²-4x²y+4xy

=x(x-4xy+y)

b)đề sai

Bài 3:

3yx + 6x - y = 7

<=> x(3y+6) - (3y+6) = 27

<=> (3y+6)(x+1) = 27

Ta có bảng sau:

Vậy...

a) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)

\(=\left[\left(x+a\right)\left(x+4a\right)\right]\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4\)

\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

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

\(=\left(x^2+5ax+5a^2\right)^2\)

b) \(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)

\(=\left(x^2+y^2+z^2\right)\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]+\left(xy+yz+zx\right)^2\)

\(=\left(x^2+y^2+z^2\right)^2+2\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)+\left(xy+yz+zx\right)^2\)

\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)

Lớp 1 thì làm gì mà có cái này hả dfdfdfdfdfd

câu b: (x^2+2x-3)/(x^2+3x-10).(x^2-9x+14)/(x^2+7x+12)

= (x^2+3x-x-3)/(x^2-2x+5x-10).(x^2-7x-2x+14)/(x^2+4x+3x+12)

=(x+3)(x-1)/(x-2)(x-5).(x-7)(x-2)/(x+4)(x+3)

=(x+3)(x-1)(x-7)(x-2)/(x-2)(x-5)(x+4)(x+3)

=(x-1)(x-7)/(x-5)(x+4)

a) 25x² - y² + 4y - 4 

= (5x)² - (y - 2)² 

= (5x + y - 2)(5x - y + 2)

b) a² + b² - x² - y² + 2ab - 2xy 

= (a² + 2ab + b²) - (x² + 2xy + y²)

= (a + b)² - (x + y)²

= (a + b + x + y)(a + b - x - y)

c) 5x²(x - 1) + 10xy(x - 1) - 5y²(1 - x)

= 5x²(x - 1) + 10xy(x - 1) + 5y²(x - 1)

= (x - 1)(5x² + 10xy + 5y²)

= 5(x - 1)(x² + 2xy + y²) 

= 5(x -1)(x  + y)²

d) x⁵ - x⁴y - xy⁴ + y⁵

= x⁴(x - y) - y⁴(x - y)

= (x - y)(x⁴ - y⁴) 

= (x - y)(x² - y²)(x² + y²) = (x - y)²(x + y)(x² + y²) 

Chu choa, đi hỏi khắp nơi luôn kìa trời!

x⁵+x⁴+x³+x²+x+1

=x³.(x²+x+1)+(x²+x+1)

=(x²+x+1)(x³+1)

=(x²+x+1)(x+1)(x²-x+1)

Bài 1: 

a: \(3x\left(x-a\right)+4a\left(a-x\right)\)

=3x(x-a)-4a(x-a)

=(x-a)(3x-4a)

b: \(x^2\left(y^2+z\right)+y^3+yz\)

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

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

c: \(3x^2\left(x+1\right)-5x\left(x+1\right)^2+4\left(x+1\right)\)

\(=\left(x+1\right)\left[3x^2-5x\left(x+1\right)+4\right]\)

\(=\left(x+1\right)\left(3x^2-5x^2-5x+4\right)\)

\(=\left(x+1\right)\left(-2x^2-5x+4\right)\)

a) x² - 7xy - 18y²

= x² + 2xy - 9xy - 18y²

= x(x + 2y) - 9y(x + 2y) 

= (x - 9y)(x + 2y) 

b) 4x² + 8x - 5

= 4x² - 2x + 10x - 5

= 2x(2x - 1) + 5(2x - 1) 

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

c) 4x⁴ - 21x²y² + y⁴

= (4x⁴ + 4x²y² + y⁴) -25x²y²

= (2x² + y²) - (5xy)²

= (2x² + 5xy + y²)(2x² - 5xy + y²) 

= \(2\left(x^2+\frac{5}{2}xy+\frac{y^2}{2}\right)2\left(x^2-\frac{5}{2}xy+\frac{y^2}{2}\right)\)

= \(4\left[\left(x+\frac{5}{4}y\right)^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\left[\left(x-\frac{5}{4}\right)y^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\)

\(=4\left(x+\frac{5}{4}y-\frac{\sqrt{17}}{4}y\right)\left(x+\frac{5}{4}y+\frac{\sqrt{17}}{4}y\right)\left(x-\frac{5}{4}y-\frac{\sqrt{17}y}{4}\right)\left(x-\frac{5}{4}y+\frac{\sqrt{17y}}{4}\right)\)

Trả lời:

a, x² - 7xy - 18y² 

= x² - 9xy + 2xy - 18y²

= ( x² - 9xy ) + ( 2xy - 18y² )

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

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

b, 4x² + 8x - 5

= 4x² + 10x - 2x - 5

= ( 4x² + 10x ) - ( 2x + 5 )

= 2x ( 2x + 5 ) - ( 2x + 5 )

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