Question

bài 1 thực hiện phép tính 

a) (3+x)^2

b) (5-x)^3

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

d) 9/x^2+3x - 3-x/x

Answer 1

a) ( 3 + x )² = x² + 6x + 9

b) ( 5 - x )³ = 125 - 75x + 45x² - x³

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

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

\(=\frac{9}{x\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

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

Answer 2

a, \(\left(3+x\right)^2=9+6x+x^2\)

b, \(\left(5-x\right)^3=125-75x+15x^2-x^3\)

c, \(\left(2x-1\right)\left(x^2-x+3\right)=2x^3-2x^2+6x-x^2+x-3=2x^3-3x^2+7x-3\)

d, \(\frac{9}{x^2+3x}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

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

Answer 3

a, \(\left(3+x\right)^2=9+6x+x^2=x^2+6x+9\)

b, \(\left(5-x\right)^3=125-75x+15x^2-x^3=x^3+15x^2-75x+125\)

c, \(\left(2x-1\right)\left(x^2-x+3\right)=2x^3-2x^2+6x-x^2+x-3=2x^3-3x^2+7x-3\)

d, \(\frac{9}{x^2+3x}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

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

Answer 4

Bổ sung nhé, thường thường thì mình cũng hay quên rút gọn x bỏ qua cho mình nhé ! 

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