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

Ta có: \(\left|x+1\right|+3\left|x-1\right|=x+2+\left|x\right|+2\left|x-2\right|\)

=>\(\left|x+1\right|-\left|x\right|+3\left|x-1\right|-2\left|x-2\right|=x+2\) (1)

TH1: x<-1

=>x+1<0; x<0; x-1<0; x-2<0

(1) sẽ trở thành:

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

=>-x-1+x-3x+3+2x-4=x+2

=>-x-2=x+2

=>-2x=4

=>x=-2(nhận)

TH2: -1<=x<0

=>x+1>=0; x<0; x-1<0; x-2<0

(1) sẽ trở thành:

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

=>x+1+x-3x+3+2x-4=x+2

=>x=x+2

=>0=2(vô lý)

TH3: 0<=x<1

=>x+1>0; x>=0; x-1<0; x-2<0

(1) sẽ trở thành:

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

=>1-3x+3+2x-4=x+2

=>-x=x+2

=>-2x=2

=>x=-1(loại)

TH4: 1<=x<2

=>x+1>0; x>0; x-1>=0; x-2<0

(1) sẽ trở thành:

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

=>1+3x-3+2x-4=x+2

=>5x-6=x+2

=>4x=8

=>x=2(loại)

TH5: x>=2

=>x+1>0; x>0; x-1>0; x-2>=0

(1) sẽ trở thành:

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

=>1+3x-3-2x+4=x+2

=>x+2=x+2(luôn đúng)

=>x>=2

Vậy: x>=2 hoặc x=-2

bạn tách từng bài ra bn

cùng 1 bài mà