Question

[Sakura서 정]

thực hiện phép tính

a) \(\left(3-2x\right).\left(x+1\right)+x.\left(2x-1\right)\)
b) \(\frac{x^2+9}{x^2+3x}+\frac{6}{x+3}\)
c) \(\frac{2+x}{2-x}+\frac{4x^2}{4-x^2}+\frac{x-2}{2+x}\)
d) \(\left(x^3+4x^2+6x+4\right):\left(x+2\right)\)

Answer 1

[bảo phạm]

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

b) \(\frac{x^2+9}{x^2+3x}+\frac{6}{x+3}=\frac{x^2+9}{x\left(x+3\right)}+\frac{6x}{x\left(x+3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x+3\right)}=\frac{x+3}{x}\)
c)\(\frac{2+x}{2-x}+\frac{4x^2}{4-x^2}+\frac{x-2}{2+x}=\frac{\left(x+2\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}+\frac{-\left(x-2\right)^2}{\left(2+x\right)\left(2-x\right)}\)
\(=\frac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}=\frac{4x^2+8x}{\left(x+2\right)\left(2-x\right)}=\frac{4x\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}=\frac{4x}{2-x}\)
d) \(\left(x^3+4x^2+6x+4\right):\left(x+2\right)\)
\(=\left(x^3+2x^2+2x^2+4x+2x+4\right):\left(x+2\right)\)
\(=\left[x^2\left(x+2\right)+2x\left(x+2\right)+2\left(x+2\right)\right]:\left(x+2\right)\)

\(=\left(x^2+2x+2\right)\left(x+2\right):\left(x+2\right)=x^2+2x+2\)

Answer 2

[Sakura서 정]

câu b sai đề r b ơi