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26 tháng 3 2023

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

26 tháng 3 2023

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)`

27 tháng 3 2020

\(\text{a) 5(2x-3)-4(5x-7)=19-2(x+11)}\)

\(10x-15-20x+28=19-2x-22\)

\(10x-20x+2x=19-22-28+15\)

\(-8x=-16\)

\(\Rightarrow x=2\)

\(\text{b) 4(x+3)-7x+17=8(5x-1)+166}\)

\(4x+12-7x+17=40x-8+166\)

\(4x-7x-40x=-8+166-17-12\)

\(-43x=129\)

\(x=-3\)

\(\text{c) 17-14(x+1)=13-4(x+1)-5(x-3)}\)

\(17-14x+14=13-4x-4-5x+15\)

\(-14x+4x+5x=13-4+15-14-17\)

\(-5x=-7\)

\(x=\frac{7}{5}\)

\(\text{d) 5x+3,5+(3x-4)=7x-3(x-0,5)}\)

\(5x+3,5+3x-4=7x-3x+1,5\)

\(5x+3x-7x+3x=1,5-3,5\)

\(x=-2\)

\(\text{e) 7(4x+3)-4(x-1)=15(x+0,75)+7}\)

\(28x+21-4x+4=15x+11,25+7\)

\(28x-4x-15x=11,25+7-4-21\)

\(9x=\frac{-27}{4}\)

\(x=\frac{-3}{4}\)

\(\text{f) 3x+2,42+o,8x=3,38-0,2x}\)

\(3x+0,8x+0,2x=3,38-2,42\)

\(4x=\frac{24}{25}\)

\(x=\frac{6}{25}\)

chúc bạn học tốt !!

7 tháng 2 2022

A 3x-4x=-9-3

    -x=-12

     x=12

B 3.2x -5x +1=5+0.2x

  3.2x-5x-0.2x=5-1

  -2x=4

 x=-2

C 1.5-x-2=-3x-0.3

  -x+3x=-0.3-1.5+2

  2x =0.2

  x=0.1

E 2/3-1/2x-1=-x+1

  -1/2x+x=1+1-2/3

  1/2x=4/3

  x=8/3

F 3t-4+13+2t+4-3t

  =3t+2t-3t-4+13+4

  =2t+13

a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)

b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)

c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

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