1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

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

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

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

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

e: =>x^3+1-2x=x^3-x

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2

a. (x - 2)(x + 2) - (x - 3)² = 9

<=> x² - 2² - (x - 3)² = 3²

<=> x - 2 - (x - 3) = 3

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

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

<=> 0 = 2 (Vô lí)

Vậy nghiệm của PT là S = \(\varnothing\)

b: Ta có: \(\left(x-1\right)\left(x^2+1\right)-\left(x+1\right)\left(x^2-x+1\right)=x\left(2-x\right)\)

\(\Leftrightarrow x^3+x-x^2-1-x^3-1=2x-x^2\)

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

\(\Leftrightarrow-x=2\)

hay x=-2

Làm hết giúp mk vs

2 nha bn

lớp 1 căng đét

\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)

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

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

\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)

\(\Leftrightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow2x+4=2x+3\)

\(\Leftrightarrow0x=-1\)(vô nghiệm)

Vậy phương trình vô nghiệm.

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow2x+7=-10\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)

Trả lời:

a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)\(\left(đkxđ:x\ne-1\right)\)

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

\(\Rightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow4+2x=2x+3\)

\(\Leftrightarrow2x-2x=3-4\)

\(\Leftrightarrow0x=-1\)(không thỏa mãn)

Vậy \(S=\varnothing\)

b, \(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\)\(\left(đkxđ:x\ne\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{\left(x+2\right)^2-\left(2x-3\right)}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow x^2+2x+7=x^2-10\)

\(\Leftrightarrow x^2+2x-x^2=-10-7\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(tm)

Vậy \(S=\left\{\frac{-17}{2}\right\}\)

c, \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)\(\left(đkxđ:x\ne1\right)\)

\(\Leftrightarrow\frac{2-5x}{2x-2}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)

\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)

\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)}+\frac{2\left(x^2+x-3\right)}{2\left(x-1\right)}\)

\(\Rightarrow2-5x+2x^2-3x+1=2x-2+2x^2+2x-6\)

\(\Leftrightarrow2x^2-8x+3=2x^2+4x-8\)

\(\Leftrightarrow2x^2-8x-2x^2-4x=-8-3\)

\(\Leftrightarrow-12x=-13\)

\(\Leftrightarrow x=\frac{13}{12}\)(tm)

Vậy \(S=\left\{\frac{13}{12}\right\}\)

1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)

ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)

<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)

<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)

<=> \(\frac{3x+10}{x^2+2x-3}=0\)

<=> \(3x+10=0\)

<=> \(x=-\frac{10}{3}\)