ĐKXĐ : \(x\notin\left\{0;-1;-2;-3;-4\right\}\)

Ta có \(\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\)

\(\Leftrightarrow\dfrac{2x+4}{x.\left(x+4\right)}+\dfrac{2x+4}{\left(x+1\right).\left(x+3\right)}+\dfrac{1}{x+2}=0\)

\(\Leftrightarrow\dfrac{2x+4}{\left(x+2\right)^2-4}+\dfrac{2x+4}{\left(x+2\right)^2-1}+\dfrac{1}{x+2}=0\) (*)

Đặt x + 2 = a \(\left(a\ne0\right)\) 

(*) \(\Leftrightarrow\dfrac{2a}{a^2-4}+\dfrac{2a}{a^2-1}+\dfrac{1}{a}=0\)

\(\Leftrightarrow\dfrac{2}{a-\dfrac{4}{a}}+\dfrac{2}{a-\dfrac{1}{a}}+\dfrac{1}{a}=0\) (**)

Đặt \(\dfrac{1}{a}=b\left(b\ne0\right)\) \(\Rightarrow ab=1\)

Ta được (**) \(\Leftrightarrow\dfrac{2}{a-4b}+\dfrac{2}{a-b}+b=0\)

\(\Leftrightarrow\dfrac{2b}{1-4b^2}+\dfrac{2b}{1-b^2}+b=0\)

\(\Leftrightarrow\dfrac{2}{1-4b^2}+\dfrac{2}{1-b^2}=-1\)

\(\Rightarrow4-10b^2=-4b^4+5b^2-1\)

\(\Leftrightarrow4b^4-15b^2+5=0\) (***)

Đặt b² = t > 0

Ta có (***) <=> \(4t^2-15t+5=0\Leftrightarrow t=\dfrac{15\pm\sqrt{145}}{8}\) (tm)

\(\Leftrightarrow b=\pm\sqrt{\dfrac{15\pm\sqrt{145}}{8}}\) 

mà x + 2 = a ; ab = 1 

nên \(x=\pm\sqrt{\dfrac{8}{15\pm\sqrt{145}}}-2\)

Thử lại ta có phương trình có 4 nghiệm như trên

\(ĐK:x\ne-1;x\ne1\\ PT\Leftrightarrow\dfrac{\dfrac{2x^2+4x+2-x^2+2x-1}{2\left(x+1\right)\left(x-1\right)}}{\dfrac{x-1+x+1}{x-1}}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2+6x+1}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2x}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2+6x+1}{4x\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow x^2+6x+1=2x\left(x-1\right)\\ \Leftrightarrow x^2+6x+1=2x^2-2x\\ \Leftrightarrow x^2-8x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{17}\left(tm\right)\\x=4-\sqrt{17}\left(tm\right)\end{matrix}\right.\)

đk : x khác 1 ; -1 

<=> \(-x\left(x+1\right)+x^2+2=2\left(x-1\right)\)

\(\Leftrightarrow-x+2=2x-2\Leftrightarrow x=\dfrac{4}{3}\)(tm)

\(\Leftrightarrow-x\left(x+1\right)+x^2+2=2x-2\)

\(\Leftrightarrow-x^2-x+x^2+2-2x+2=0\)

=>-3x+4=0

hay x=4/3(nhận)

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

\(\Leftrightarrow-x^2-x+x^2+2-2x+2=0\left(quy\cdotđồng\cdot và\cdot khử\cdot mẫu\right)\)

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

\(\Leftrightarrow-3x=-4\)

\(\Leftrightarrow x=\dfrac{-4}{3}\)

Vậy \(S=\left\{-\dfrac{4}{3}\right\}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x-1}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x-1+x-2}{\left(x-1\right)\left(x-2\right)}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{2x-3}{x^2-3x+2}>\dfrac{1}{x}\\ \Leftrightarrow x\left(2x-3\right)>x^2-3x+2\\ \Leftrightarrow2x^2-3x>x^2-3x+2\\ \Leftrightarrow x^2>2\\ \Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)

TK

https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5

a: \(\Leftrightarrow4x-5=2x-2+x\)

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

hay x=41/4(nhận)

c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

=>-6x+16=-5x+11

=>-x=-5

hay x=5(nhận)

d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)

\(\Leftrightarrow4x=16\)

hay x=4(nhận)

đk : x khác -1 ; 1 

\(1+5x-5-x-1=x^2-1\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)(tm) 

ĐKXĐ:\(x\ne-1,x\ne0\)

\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}-\dfrac{2x+1}{x\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-1+x-2x-1}{x\left(x+1\right)}=0\\ \Rightarrow x^2-x-2=0\\ \Leftrightarrow x^2-2x+x-2=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

Vậy pt có tập nghiệm `S={2}`

\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\left(đk:x\ne0,-1\right)\)

\(\Leftrightarrow\dfrac{x-1}{x}+\dfrac{1}{x+1}-\dfrac{2x+1}{x\left(x+1\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)+x-2x-1}{x\left(x+1\right)}=0\)

\(\Leftrightarrow x^2+x-x-1+x-2x-1=0\)

\(\Leftrightarrow x^2-x-2=0\)

\(\Delta=b^2-4ac=\left(-1\right)^2-4.\left(-2\right)=9>0\Rightarrow\sqrt{\Delta}=3\)

\(\Rightarrow\)PT có 2 nghiệm \(x_1,x_2\)

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+3}{2}=2\left(n\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-3}{2}=-1\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

x = 2

ĐKXĐ: `{(x+1>0),(x ne0):} <=> {(x> -1),(x ne 0):}`

`2/(sqrt(x+1))+1/(x sqrt(x+1)) =1/x`

`<=>(2x+1)/(x sqrt(x+1)) =1/x`

`<=>x(2x+1)=x sqrt(x+1)`

`<=>2x+1=sqrt(x+1)`

`=>(2x+1)^2=x+1`

`<=>4x^2+4x+1=x+1`

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

`<=>x(4x+3)=0`

`<=>[(x=0\ (KTM)),(x=-3/4):}`

Thay `x=-3/4` vào PT ban đầu `=>` Không thỏa mãn.

Vậy phương trình vô nghiệm.

a: =>10x=3(5-3x)

=>10x=15-9x

=>19x=15

=>x=15/19

b: =>\(\dfrac{x\left(x-4\right)+x^2-1}{x\left(x+1\right)}=2\)

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

=>2x=-4x-1

=>6x=-1

=>x=-1/6

c:=>x(x+2)-x+2=2

=>x^2+2x-x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

d: =>x+1+3x=2

=>4x=1

=>x=1/4

e: =>x(x+1)+x(x-3)=2x

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

=>2x^2-4x=0

=>x=0(nhận) hoặc x=2(nhận)

f: =>2x+6-4x+12=5

=>-2x=-13

=>x=13/2