a) \(\dfrac{5x}{2x+2}+1=\dfrac{6}{x+1}\left(đk:x\ne-1\right)\)

\(\dfrac{5x+2x+2}{2x+2}=\dfrac{12}{2x+2}\)

\(7x+2=12\)

\(7x=10\)

\(x=\dfrac{10}{7}\left(TM\right)\)

b) \(\dfrac{-48}{x^2-9}=\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}\left(đk:x\ne\pm3\right)\)

\(\left(x-3\right)^2-\left(x+3\right)^2=-48\)

\(x^2-6x+9-x^2-6x-9=-48\)

\(x^2-12x+48=0\)

\(\left(x-6\right)^2=-12\)

Vì \(\left(x-6\right)^2\ge0\forall x\)

\(\Rightarrow\) pt vô nghiệm

a) ĐKXĐ: x≠-5

Ta có: \(\dfrac{2x-5}{x+5}=4\)

\(\Leftrightarrow2x-5=4\left(x+5\right)\)

\(\Leftrightarrow2x-5=4x+20\)

\(\Leftrightarrow2x-5-4x-20=0\)

\(\Leftrightarrow-2x-25=0\)

\(\Leftrightarrow-2x=25\)

hay \(x=\dfrac{-25}{2}\)(nhận)

Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)

b) ĐKXĐ: x≠0

Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)

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

\(\Leftrightarrow2x^2-8-2x^2-3x=0\)

\(\Leftrightarrow-3x-8=0\)

\(\Leftrightarrow-3x=8\)

hay \(x=\dfrac{-8}{3}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)

Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)

\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)

\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)

\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)

\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)

\(\Leftrightarrow20x+12=0\)

\(\Leftrightarrow20x=-12\)

hay \(x=-\dfrac{3}{5}\)(nhận)

Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)

d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)

\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

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

Ta có: \(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)

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

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

\(\Leftrightarrow\dfrac{5x+5-2x+4}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-3}{2x-1}\)

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

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

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

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

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

\(\Leftrightarrow x^2+x-3=0\)

\(\Delta=1^2-4\cdot1\cdot\left(-3\right)=13\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{13}}{2}\left(nhận\right)\\x_2=\dfrac{-1+\sqrt{13}}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{-1-\sqrt{13}}{2};\dfrac{-1+\sqrt{13}}{2}\right\}\)

Mk giải giúp bạn phần a thôi nha! (Dài lắm, lười :v)

a, 1 + \(\dfrac{x}{3-x}\) = \(\dfrac{5x}{\left(x+2\right)\left(x+3\right)}+\dfrac{2}{x+2}\) (x \(\ne\) -2; x \(\ne\) \(\pm\) 3)

\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{5x+2\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}\)

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

\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{7x+6}{x^2+5x+6}\)

Vì 3 - x \(\ne\) 0; x² + 5x + 6 \(\ne\) 0

\(\Rightarrow\) 3(x² + 5x + 6) = (7x + 6)(3 - x)

\(\Leftrightarrow\) 3x² + 15x + 18 = 21x - 7x² + 18 - 6x

\(\Leftrightarrow\) 10x² = 0

\(\Leftrightarrow\) x = 0 (TM)

Vậy S = {0}

Chúc bn học tốt! (Nếu bạn cần phần nào khác mk có thể giúp bn chứ đừng có đăng hết lên, ít người làm lắm :v)

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

vậy..

d)\(\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(2-x\right)}+\dfrac{\left(x+3\right)\left(x-3\right)}{\left(2-x\right)\left(x-3\right)}=0\\ \Leftrightarrow5+x^2-9=0\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

vậy..

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)

a) ĐKXD: x ≠ 2

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

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{3-x}{x-2}=-3\)

\(\Leftrightarrow\dfrac{1-3+x}{x-2}=-3\)

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

\(\Leftrightarrow-2+x=-3\left(x-2\right)\)

\(\Leftrightarrow-2+x=-3x+6\)

\(\Leftrightarrow x+3x=6+2\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\) (loại vì không thỏa mãn điều kiện)

Vậy S = ∅

b) ĐKXĐ: x ≠ 7

 \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

\(\Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{1}{x-7}=8\)

\(\Leftrightarrow\dfrac{7-x}{x-7}=8\)

\(\Leftrightarrow-1=8\left(vô-lý\right)\)

Vậy S = ∅ 

P/s: Ko chắc ạ! 

c) ĐKXĐ: x ≠ 1

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

Quy đồng và khử mẫu ta được:

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

\(\Leftrightarrow x^2+x+1+2x^2-2x-3x^2=0\)

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

\(\Leftrightarrow x=1\) (loại vì ko t/m đk)

Vậy S = ∅

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

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

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

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

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

$\Leftrightarrow (x-1)^2=2^2$

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

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

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

$\Leftrightarrow (x^2-2x)-(3x-6)=0$

$\Leftrightarrow x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

2f.

ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$

$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$

$\Leftrightarrow x^2-4x+4-3x-6=2x-22$

$\Leftrightarrow x^2-7x-2=2x-22$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Leftrightarrow x-4=0$ hoặc $x-5=0$

$\Leftrightarrow x=4$ hoặc $x=5$ (tm)

`a,x(x+3)-(2x-1).(x+30)=0`
`<=>x^2+3x-(2x^2+59x-30)=0`
`<=>x^2+56x-30=0`
`<=>x^2+56x+28^2=28^2+30`
`<=>(x+28)^2=28^2+30`
`<=>x=+-sqrt{28^2+30}-28`
`b,x(x-3)-5(x-3)=0`
`<=>(x-3)(x-5)=0`
`<=>` $\left[ \begin{array}{l}x=3\\x=5\end{array} \right.$
`c)1/(x-1)+5/(x-2)=(3x)/((x-1)(x-2))`
`đk:x ne 1,2`
`pt<=>x-2+5(x-1)=3x`
`<=>x-2+5x-5=3x`
`<=>6x-7=3x`
`<=>3x=7`
`<=>x=7/3`
`d)(x-1)/(x+1)+(x+1)/(x-1)=(4-2x^2)/(x^2-1)`
`đk:x ne +-1`
`pt<=>(x-1)^2+(x+1)^2=4-2x^2`
`<=>2x^2+2=4-2x^2`
`<=>4x^2=2`
`<=>x^2=1/2`
`<=>x=+-sqrt{1/2}`

a: \(\Leftrightarrow3x+2\left(x+2\right)=5\left(x-1\right)\)

=>3x+2x+4=5x-5

=>4=-5(vô lý)

b: \(\Leftrightarrow\dfrac{2}{x\left(x+4\right)}-\dfrac{3x}{x+4}=-3\)

\(\Leftrightarrow2-3x^2=-3x\left(x+4\right)\)

\(\Leftrightarrow2-3x^2+3x^2+12x=0\)

=>12x+2=0

hay x=-1/6