8(x-1)-4=6(x+2)-2

8x-8-4=6x+12-2

8x-6x=12-2+8+4

2x=22

x=11

<=> 8x-8-4=6x+12-12

<=> 8x-6x=12-12+8+4

<=> 2x=12

<=> x=6

\(g'=2\left(\sqrt{x+3}\right)^2.\left(\sqrt{x+3}\right)'=2\left(x+3\right).\dfrac{1}{2\sqrt{x+3}}=\sqrt{x+3}\)

\(g'\left(x\right)+\sqrt{2x-1}=3\Leftrightarrow\sqrt{x+3}+\sqrt{2x-1}=3\)

\(DKXD:x\ge\dfrac{1}{2}\)

\(pt\Leftrightarrow x+3+2x-1+2\sqrt{\left(x+3\right)\left(2x-1\right)}=9\)

\(\Leftrightarrow2\sqrt{\left(x+3\right)\left(2x-1\right)}=7-3x\)

\(\Leftrightarrow4\left(2x^2+5x-3\right)=49-42x+9x^2\)

\(\Leftrightarrow x^2-62x+61=0\Leftrightarrow\left[{}\begin{matrix}x=61\left(loai\right)\\x=1\end{matrix}\right.\)

g^'(x) = \(\sqrt{x+3}\) 

ta có phương trình : \(\sqrt{x+3}\)  + \(\sqrt{2x-1}\) =3 ( ĐK : x\(\ge\)\(\dfrac{1}{2}\))

\(\Leftrightarrow\) x+3 +2x-1 +\(2\sqrt{\left(x+3\right)\left(2x-1\right)}\) = 9

\(\Leftrightarrow\) \(2\sqrt{\left(x+3\right)\left(2x-1\right)}\) = 7-3x

\(\Leftrightarrow\) 4(2x² +5x -3) = 49 - 42x +9x² 

\(\Leftrightarrow\) x² - 62x +61 = 0 \(\left\{{}\begin{matrix}x=61\\x=1\end{matrix}\right.\)

\(\dfrac{3}{1-x}-\dfrac{2}{x+2}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\left(x\ne1;x\ne-2\right)\)

\(< =>\dfrac{-3}{x-1}-\dfrac{2}{x+2}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\)

\(< =>\dfrac{-3\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{2\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\)

suy ra

`-3(x+2)-2(x-1)=x+8`

`<=>-3x-6-2x+2=x+8`

`<=>-3x-2x-x=8+6-2`

`<=>-6x=12`

`<=>x=-2(ktmđk)`

Vậy phương trình vô nghiệm

=>-3(x+2)-2x+2=x+8

=>-3x-6-2x+2=x+8

=>-5x-4=x+8

=>-6x=12

=>x=-2(loại)

Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?

Câu này tớ giải hơn 10 lần rồi cậu ( ko xàm :) 

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

Do đó \(x+100=0\Leftrightarrow x=-100\)

Vậy pt có nghiệm : x=-100

Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

=> \(\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

=> \(\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

=> \(\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

=> x + 100 = 0

=> x = - 100

Vậy x = - 100

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{98}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}\right)=\left(x+100\right)\left(\frac{1}{94}+\frac{1}{92}\right)\)

Do \(\frac{1}{98}+\frac{1}{96}\ne\frac{1}{94}+\frac{1}{92}\)\(\Rightarrow x+100=0\)

                                                    \(\Leftrightarrow x=-100\)

           Vậy \(x=-100\)

~Hok Tốt~

bạn tách từng bài ra bn

tl câu hỏi trên cho mik ik