\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

4 câu đầu hìn như sai đề :v

`m)(3/2-2/(-5)):x-1/2=3/2`

`<=>(3/2+2/5):x=3/2+1/2=2`

`<=>19/10:x=2`

`<=>x=19/10:2=19/20`

`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`

`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`

`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`

`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`

Mà `3/2-5/11-3/13>0`

`<=>2x-2+1/2=0`

`<=>2x-3/2=0`

`<=>2x=3/2<=>x=3/4`

h, \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\left(x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2}{2}-1=\dfrac{x}{12}\)

\(\Leftrightarrow x^2-\dfrac{x}{6}-2=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{12}+\dfrac{1}{144}-\dfrac{289}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{12}\right)^2=\dfrac{289}{144}\)

\(\Leftrightarrow x=\dfrac{1}{12}\pm\dfrac{\sqrt{289}}{12}\)

Vậy ...

i, \(\Leftrightarrow x^2-\dfrac{2.x.7}{12}+\dfrac{49}{144}-\dfrac{1}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{144}\)

\(\Leftrightarrow x=\dfrac{7}{2}\pm\dfrac{1}{12}\)

Vậy ...

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12x^2-24-2x=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-8}{3}\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{8}{3};\dfrac{3}{2}\right\}\)

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

\(\Leftrightarrow\dfrac{19}{10}:x=2\)

hay \(x=\dfrac{19}{20}\)

Vậy: \(S=\left\{\dfrac{19}{20}\right\}\)

1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)

Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)

\(\Leftrightarrow-3x-12-3+5x-x+4=0\)

\(\Leftrightarrow x=11\left(nhận\right)\)

2. ĐKXĐ: $x\neq \pm 2$

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

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

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

\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)

Vậy pt vô nghiệm

3. ĐKXĐ: $x\neq \pm \frac{3}{2}$

PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)

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

\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)

\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)

a: ĐKXĐ: \(x\notin\left\{2;5\right\}\)

\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)

=>6x+1+5x-25-3x+6=0

=>8x-18=0

=>8x=18

=>\(x=\dfrac{9}{4}\left(nhận\right)\)

b: Đề thiếu vế phải rồi bạn

c: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

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

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

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

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

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

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

=>5x=3

=>\(x=\dfrac{3}{5}\left(nhận\right)\)

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

Suy ra: \(3x-6-x+1=2x+4\)

\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)

2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)

Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)

\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)

\(\Leftrightarrow5x=16\)

hay \(x=\dfrac{16}{5}\)

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: \(S=\left\{-\dfrac{1}{56}\right\}\)

\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)

\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)

g.\(\dfrac{1-3x}{6}+x-1=\dfrac{x+2}{2}\)

\(\Leftrightarrow\dfrac{\left(1-3x\right)+6\left(x-1\right)}{6}=\dfrac{3\left(x+2\right)}{6}\)

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

\(\Leftrightarrow1-3x+6x-6=3x+6\)

\(\Leftrightarrow-5=6\left(vô.lí\right)\)

Vậy pt vô nghiệm

h.\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)

\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)

\(\Leftrightarrow30x+15-100-6x-4=24x-8\)

\(\Leftrightarrow-89=-8\left(vô.lí\right)\)

Vậy pt vô nghiệm

i.\(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)

\(\Leftrightarrow\dfrac{21\left(4x+3\right)-15\left(6x-2\right)}{105}=\dfrac{35\left(5x+4\right)+215}{105}\)

\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+215\)

\(\Leftrightarrow84x+63-90x+30=175x+140+215\)

\(\Leftrightarrow-181=262\)

\(\Leftrightarrow x=-\dfrac{262}{181}\)

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)