Question

thực hiện phép tính

(x+1)^2 + (x-2) (x+3) - 4x

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

Answer 1

Trả lời:

a, ( x + 1 )² + ( x - 2 ) ( x + 3 ) - 4x 

= x² + 2x + 1 + x² + 3x - 2x - 6 - 4x

= 2x² - x - 5

b, ( x - 2 )² + ( x + 1 )² + 2 ( x - 2 ) ( - 1 - x ) 

= x² - 4x - 4 + x² + 2x + 1 + ( 2x - 4 ) ( - 1 - x )

= 2x² - 2x - 3 - 2x - 2x² + 4x + 4x

= 4x - 3

Answer 2

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

\(=\left(x^2+2x+1\right)+\left(x^2+x-6\right)-4x\)

\(=x^2+2x+1+x^2+x-6-4x\)

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

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

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

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

\(=x^2-4x+4+x^2+2x+1-2x-2x^2+4+4x\)

\(=9\)