Tử mẫu không rõ rằng => lạc đề

`2)x^4+2x^3-x^2-2x+1=0`

`<=>x^4+2x^3+x^2-2x^2-2x+1=0`

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

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

`<=>x^2+x-1=0`

`\Delta=1+4=5`

`=>x_{1,2}=(-1+-sqrt5)/2`

Vậy `S={(-1+sqrt5)/2,(-1+sqrt5)/2`

`3)x^4-4x^3-9x^2+8x+4=0`

`<=>x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0`

`<=>(x-1)(x^3-3x^2-12x-4)=0`

`<=>(x-1)(x^3+2x^2-5x^2-10x-2x-4)=0`

`<=>(x-1)(x+2)(x^2-5x-10)=0`

`+)x=1`

`+)x=-2`

`+)x^2-5x-10=0`

`Delta=25+40=65`

`=>x_{12}=(5+sqrt{65})/2`

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)

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

d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)

Tương tự với a ; b 

\(\frac{4}{2x-3}-\frac{1}{2x+3}+\frac{2x+9}{9-4x^2}\)

\(\Leftrightarrow\frac{4}{2x-3}-\frac{1}{2x+3}+\frac{-2x-9}{4x^2-9}\)

\(\Leftrightarrow\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}+\frac{-2x-9}{\left(2x+3\right)\left(2x-3\right)}\)

\(\Leftrightarrow\frac{8x+12-2x+3+2x-2x-9}{\left(2x-3\right)\left(2x+3\right)}\)

\(\Leftrightarrow\frac{6x+6}{\left(2x-3\right)\left(2x+3\right)}\)

\(\Leftrightarrow\frac{2\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}\)

\(\Leftrightarrow\frac{2}{2x-3}\)

xem câu hỏi mình vừa làm Câu hỏi của dungdt0112 - Toán lớp 8 - Học toán với OnlineMath