Question

b3:
a)l 2-3x l =-1
b)3x-2,42+0,8=3,38-0,2x
c)3/x-1 + 2/x²+x+1 = 3x²/x³-1

Answer 1

đề bài yêu cầu là "giải các phương trình sau " ạ

Answer 2

a)

\(\left|2-3x\right|=-1\) (vô lí vì \(\left|2-3x\right|\ge0\) )

b)

`3x-2,42+0,8=3,38-0,2x`

`<=>3x+0,2x=3,38+2,42-0,8`

`<=>3,2x=5`

`<=>x=25/16`

c)

\(\dfrac{3}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\left(x\ne1\right)\)

\(< =>\dfrac{3}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(< =>\dfrac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)

suy ra

`3x^2 +3x+3+2x-2=3x^2`

`<=>3x^2 -3x^2 +3x+2x=-3+2`

`<=>5x=-1`

`<=>x=-1/5(tmđk)`