a. 2x(x + y) - y(y + 2x) = 2x² + 2xy - y² - 2xy = 2x² - y²

b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

Phần c nản quá.

a) 2x(x + y) - y(y + 2x) 

= 2x² + 2xy - y² - 2xy

= 2x² - y²

b) \(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

c) \(\frac{x^3-4x^2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{x-1}\)

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

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

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

( x - y )³ - x² + 2xy - y²

= ( x - y )³ - ( x² - 2xy + y² )

= ( x - y )³ - ( x - y )²

Với x - y = 5 => Giá trị biểu thức = 5³ - 5² = 100

2x² + x = 0

<=> x( 2x + 1 ) = 0

<=> x = 0 hoặc 2x + 1 = 0

<=> x = 0 hoặc x = -1/2

b) x² + 4x + 3 = 0

<=> ( x² + 4x + 4 ) - 1 = 0

<=> ( x + 2 )² - 1² = 0

<=> ( x + 2 - 1 )( x + 2 + 1 ) = 0

<=> ( x + 1 )( x + 3 ) = 0

<=> x + 1 = 0 hoặc x + 3 = 0

<=> x = -1 hoặc x = -3

=3x2 .5x2-3x2..2x-3x.4
 

Bài làm

\(3x^2\left(5x^2-2x-4\right)=15x^4-6x^3-12x^2\)

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